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if x represents a number then write an expression for one half the sum of x and 7

translate the statement into an algebraic expression

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Writing Algebraic Expressions


Writing Algebraic Expressions is presnted by Math Goodies. ... then the expression 9 + x has a value of 11. ... each algebraic expression consisted of one number, ...
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If x represents a number, then how do you write expression ...


SOCRATIC ... then how do you write expression for one half the sum of x ... How do you write each phrase as a variable expression: the sum of twice a number n ...
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If represents a number, then write an expression for: a ...


... then write an expression for: a number that is ... If x represents a number, then write an expression for one half ... Find two consecutive integers whose sum is ...
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If z represents a number, then how do you write expression ...


... z + 7 Then to take one half (1/2) of this would be: (z + 7)/2. SOCRATIC ... Socratic Meta ... then how do you write expression for one half the sum of x and 7?
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Writing Expressions (solutions, examples, videos)


5x + 6 . 4 times the sum of a number and 7 . 4 ... and that expression and then add 6. 2: First consider the expression for ... Half of a number is 16. Write an ...
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1 - Translating English to Algebra.notebook


If x represents a number then write an expression for a number that is three ... expression for one half the sum of x and 7. ... Translating English to Algebra ...
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English Terms to Algebraic Expressions | Breakthru


Algebraic Expressions: Ten more than x: ... The product of two times a number is 10: 2x = 10: One half a number is 10: x/2 = 10: Five times the sum of x and 2: 5 ...
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HAPPY MONDAY! DIN: IF K REPRESENTS A NUMBER, THEN WRITE AN ...


... then write an expression for one half the sum of k and 7. 9/28/15. ... if k represents a number, then write an expression for one half the sum of k and 7. 9/28/15 ...
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The sum of two numbers is 12. One number is x. The other ...


The sum of two numbers is 12. One number is x. ... number, sum, total, ... x. This time the expression is the result of solving an equation.
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Suggested Questions And Answer :


if x represents a number then write an expression for one half the sum of x and 7

???????????????? "represent a number" ????????? me did NOT vote for him, so him dont represent me y=(x+7)/2
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how do I write an algebraic expression for fourth grade pre-algebra?

1.) The sum lets you know its addition, a number could be anything we don't know and 8.   we will let x represent any number ( you can use any expression)  x+8 2.)  5 < x     3.)  product lets you know its multiplication, again one is unknown and the other is 9. you can write this like 9x or 9(x) or 9 x X 4.) this is the same as the one above, 4x or 4 x X (IF YOU DONT LIKE THE LETTER X YOU CAN USE ANY LETTER IT DOESNT MATTER like 4 x b) 5.) quotient means division: 18 divided by x or 18/x
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how to factorise fully

I can offer tips: Look out for constants and coefficients that are multiples of the same number, e.g., if all the coefficients are even, 2 is a factor. If 3 goes into all the coefficients, 3 is a factor. Place the common numerical factor outside brackets, containing the expression, having divided by the common factor. In 2x^2+10x-6 take out 2 to give 2(x^2+5x-3). In 24x^2-60x+36 12 is a common factor so this becomes: 12(2x^2-5x+3). This also applies to equations: 10x+6y=16 becomes 2(5x+3y)=2(8). In the case of an equation, the common factor can be completely removed: 5x+3y=8. Now look for single variable factors. For example, x^2y^3+x^3y^2. Look at x first. We have x^2 and x^3 and they have the common factor x^2, so we get x^2(y^3+xy^3) because x^2 times x is x^3. Now look at y. We have y^2 and y^3 so now the expression becomes x^2y^2(y+x). If the expression had been 2x^2y^3+6x^3y^2, we also have a coefficient with a common factor so we would get: 2x^2y^2(y+3x). If you break down the factors one at a time instead of all at once you won't get confused. After single factors like numbers and single variables we come to binomial factors (two components). These will usually consist of a variable and a constant or another variable, such as x-1, 2x+3, x-2y, etc. The two components are separated by plus or minus. In (2) above we had a binomial component y+x and y+3x. These are factors. It's not as easy to spot them but there are various tricks you can use to help you find them. More often than not you would be asked to factorise a quadratic expression, where the solution would be the product of two binomial factors. Let's start with two such factors and see what happens when we multiply them. Take (x-1) and (x+3). Multiplication gives x(x+3)-(x+3)=x^2+3x-x-3=x^2+2x-3. We can see that the middle term (the x term) is the result of 3-1, where 3 is the number on the second factor and 1 the number in the first factor. The constant term 3 is the result of multiplying the numbers on the factors. Let's pick a quadratic this time and work backwards to its factors; in other words factorise the quadratic. x^2+2x-48. The constant term 48 is the product of the numbers in the factors, and 2 is the difference between those numbers. Now, let's look at a different quadratic: x^2-11x+10. Again the constant 10 is the product of the numbers in the factors, but 11 is the sum of the numbers this time, not the difference. How do we know whether to use the sum or difference? We look at the sign of the constant. We have +10, so the plus tells us to add the numbers (in this case, 10+1). When the sign is minus we use the difference. So in the example of -48 we know that the product is 48 and the difference is 2. The two numbers we need are 6 and 8 because their product is 48 and difference is 2. It couldn't be 12 and 4, for example, because the difference is 8. What about the signs in the factors? We look at the sign of the middle term. If the sign in front of the constant is plus, then the signs in front of the numbers in the factors are either both positive or both negative. If the sign in front of the constant is negative then the sign in the middle term could be plus or minus, but we know that the signs within each factor are going to be different, one will be plus the other minus. The sign in the middle term tells us to use the same sign in front of the larger of the two numbers in the factors. So for x^2+2x-48, the numbers are 6 and 8 and the sign in front of the larger number 8 is the same as +2x, a plus. The factors are (x+8)(x-6). If it had been -2x the factors would have been (x-8)(x+6). Let's look at a more complicated quadratic: 6x^2+5x-21. (You may also see quadratics like 6x^2+5xy-21y^2, which is dealt with in the same way.) The way to approach this type of problem is to look at the factors of the first and last terms. Just take the numbers 6 and 21 and write down their factors as pairs of numbers: 6=(1,6), (2,3) and 21=(1,21), (3,7), (7,3), (21,1). Note that I haven't included (6,1) and (3,2) as pairs of factors for 6. You'll see why in a minute. Now we make a table (see (5) below). In this table the columns A, B, C and D are the factors of 6 (A times C) and 21 (B times D). The table contains all possible arrangements of factors. Column AD is the product of the "outside factors" in columns A and D and column BC is the product of the "inside factors" B and C. The last column depends on the sign in front of the constant 21. The "twiddles" symbol (~) means positive difference if the sign is minus, and the sum if the sign in front of the constant is plus. So in our example we have -21 so twiddles means the difference, not the sum. Therefore in the table we subtract the smaller number in the columns AD and BC from the larger and write the result in the twiddles column. Now we look at the coefficient of the middle term of the quadratic, which is 5 and we look down the twiddles column for 5. We can see it in row 6 of the figures: 2 3 3 7 are the values of A, B, C and D. If the number hadn't been there we've either missed some factors, or there aren't any (the factors may be irrational). We can now write the factors leaving out the operators that join the binomial operands: (2x 3)(3x 7). One of the signs will be plus and the other minus. Which one is which? The sign of the middle term on our example is plus. We look at the AD and BC numbers and associate the sign with the larger product. We are interested in the signs between A and B and C and D. In this case plus associates with AD because 14 is bigger than 9. The plus sign goes in front of the right-hand operand D and the minus sign in front of right-hand operand B. If the sign had been minus (-5x), minus would have gone in front of D and plus in front of B. So that's it: [(Ax-B)(Cx+D)=](2x-3)(3x+7). (The solution to 6x^2+5xy-21y^2 is similar: (2x-3y)(3x+7y).) [In cases where the coefficient of the middle term of the quadratic appears more than once, as in rows 1 and 7, where 15 is in the twiddles column, then it's correct to pick either of them, because it just means that one of the binomial factors can be factorised further as in (1) above.] Quadratic factors A B C D AD BC AD~BC 1 1 6 21 21 6 15 1 3 6 7 7 18 11 1 7 6 3 3 42 39 1 21 6 1 1 126 125 2 1 3 21 42 3 39 2 3 3 7 14 9 5 2 7 3 3 6 21 15 2 21 3 1 2 63 61  
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find the center of mass of the solid of uniform density biunded by the graphs of the equations

x^2+y^2=a^2 is a cylinder with its circular cross-section in the x-y plane; the cylinder is sliced at an angle by the plane z=cy, forming a wedge. y=0 is the x-z plane and z=0 is the x-y plane; these planes and the conditions that y and z are positive impose further constraints on the shape. The cylinder is sliced in half by the x-z plane so the cross-section will be a semicircle and only the upper half is required. The semicircular end rests on the z=0 plane, and doesn't extend beyond the vertex of the wedge. The length of the wedge is ac, because the slope of the angle of the wedge is tan^-1(c) to the y axis and the radius of the wedge is a. Viewed from the side, along the x axis, the wedge is a right-angled triangle with sides a (height), ac (length) and hypotenuse a√(1+c^2). In everyday terms, the wedge looks like the end of a lipstick that has been sliced across. Going back to the side view of the wedge, we can see that the flat end has a height a. As we move along the wedge towards the point the height changes and shrinks to zero when we reach the vertex. The length at a point P(y,z) is a-y. This height is the distance of a chord to the circumference, starting at height a. The first task is to work out the area of a segment. We do this by subtracting the area of a triangle from the area of a sector, given the distance of the chord from the circumference. This distance is a-y, so the distance from the centre of the circle is, conveniently, y. y/a=cosø where ø is half the angle subtended by the arc of the sector. So ø=cos^-1(y/a) and the area of the sector is given by 2ø/2π=ø/π=A/πa^2, or A=a^2ø (ø is measured in radians). The area of the isosceles triangle formed by joining the radial ends of the arc of the sector is B=aysinø=y√(a^2-y^2), and the area of the segment is A-B=a^2cos^-1(y/a)-y√(a^2-y^2). (Note that when y=0, this expression becomes πa^2/2, the area of the semicircle.) The volume of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz where infinitesimal dz is the thickness of the segment. The mass of the segment is directly proportional to its volume because the density is uniform, so the volume is sufficient to represent the mass. The centre of gravity (COG) of the wedge can be found by summing moments about a point G(x,y,z) which is the COG. This sum has to be zero. Since a semicircle is symmetrical in the x-y plane about the y axis, we know the x coord of G must be zero. We need G(0,g,h) where g is the height along the z axis of the COG and h the z position. We need to find the COG of a segment first. The length of the chord of a segment is 2√(a^2-y^2). Earlier we discovered this when we were finding out the areas of the sector and isosceles triangle. Consider just two dimensions. The length of the chord is one dimension and if we create a rectangle from this length and an infinitesimal width dy, the area will be 2√(a^2-y^2)dy. If we think of the rectangle as having mass, its mass is proportional to its area, so area is a sufficient measure for mass. Let's call the COG of the segment point C(x,y)=(0,q), then the moment of the rectangle about the COG=C(0,q) is mass times distance from C=2(q-y)√(a^2-y^2)dy. If we sum the rectangles over the interval y to a we have 2∫((q-y)√(a^2-y^2)dy)=0 for y≤y≤a. The interval looks strange, but it means that y is just a general value, which is determined when we return to the wedge. What we are trying to do is to find q relative to y. We need to evaluate the definite integral, so we split it: 2q∫(√(a^2-y^2)dy)-2∫(y√(a^2-y^2)dy. Call these (a) and (b). To evaluate (a), let y=asinu, then dy=acosudu; √(a^2-y^2)=acosu and the integral becomes 2a^2q∫(cos^2(u)du). Since cos(2u)=2cos^2(u)-1, cos^2(u)=½(cos(2u)+1), so we have a^2q∫((cos(2u)+1)du)=a^2q(sin(2u)/2+u). Since sin(2u)=2sinucosu, this becomes a^2q(sinucosu+u)=a^2q(y/a * √(1-(y/a)^2)+sin^-1(y/a))=qy√(a^2-y^2)+a^2qsin^-1(y/a). Now we apply the limits for y: πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a). To evaluate (b), let u=a^2-y^2, then du=-2ydy and the integral is +∫√udu = (2/3)u^(3/2)=(2/3)(a^2-y^2)^(3/2). Applying the limits we have: -(2/3)(a^2-y^2)^(3/2). (a)+(b)=πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a)-(2/3)(a^2-y^2)^(3/2)=0. Multiply through by 6: 3πa^2q-6qy√(a^2-y^2)+6a^2qsin^-1(y/a)-4(a^2-y^2)^(3/2)=0. From this q=4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a)). Fearsome though this looks, when y=0 and z=0, q=4a^3/(3πa^2)=4a/(3π), which is in fact the COG of a semicircle. When y=a and z=ac, q=0. Strictly speaking, we should write q(y) instead of just q, showing q to be a dependent variable rather than a constant. To find the COG of the wedge, we need to find the moments of the COGs of the segments. For a segment distance y above and cy along the z axis, we have the above expression for q. So q(y) is the y coord of the COG of the segment and cy is the z coord. The x coord is 0. The mass of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz, as we saw earlier, and this mass can be considered to be concentrated or focussed at the COG of the segment (0,q(y),cy). The COG of the wedge is at G(0,g,h) so the distance of the mass of the segment from G is g-q(y) and h-cy in the y and z directions. The y-moment is (g-4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a))(a^2cos^-1(y/a)-y√(a^2-y^2))dz and the z-moment is (h-cy)(a^2cos^-1(y/a)-y√(a^2-y^2))dz. The sums of these individual components are zero. These lead to rather complicated integrals, requiring considerably more space than is available to work out, even if I knew how to do so!
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write a boolean algebraic expression that represent the multiplication of two one bit numbers.

which phrase represents the algebraic expression three plus x over five? five divided by a number, increased by threethe sum of three and a number divided by fivefive more than the quotient of a number and three
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If b represents a whole number, write an expression for the sum of the next two whole numbers

b represents a whole number The next two whole number will be b+1 and b+2 Their sum will be 2b+3
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how many 1/2" candies will fit in a 41/2" x 4" bowl?

Assuming that 4" is the height of the bowl, we can conclude that it is ellipsoidal, like a rugby football, with the shorter axis (x axis) being 4.5" in diameter and its longer axis (y axis) needing to be determined from the given figures. Two domes are cut off each end of the ellipsoid leaving a 3.5" diameter circular top and a 3" diameter circular base. The general equation of the ellipse that will be used to model the bowl is x^2/2.25^2+y^2/a^2 (2.25" being the "radius" of the x axis, and a the radius of the y axis). The ellipse has its centre at the x-y origin. To find a we work out in terms of a where the base of the top dome has a width of 3.5" and where the base of the bottom dome has a width of 3". Putting x=1.75 (half of 3.5) in the equation of the ellipse, we get y=sqrt(a^2(1-1.75^2/2.25^2))=4asqrt(2)/9. Similarly for the base dome: y=asqrt(1-1.5^2/2.25^2)=-asqrt(5)/3, negative because it's below the centre or origin. The difference between these two y values is the height of the bowl, 4", so 4asqrt(2)/9+asqrt(5)/3=4, and a=36/(4sqrt(2)+3sqrt(5))=2.9114 approx.  The volume of the bowl is found by considering thin discs of radius x and thickness dy so that we can use the integral of (pi)x^2dy (the volume of a disc) between the y limits imposed by the heights of the two domes. x^2=5.0625(1-y^2/a^2). Because a is a complicated expression, we'll just use the symbol for it. We can also write 5.0625 as 81/16. The integrand becomes (81(pi)/16)(1-y^2/a^2)dy, with limits y=-asqrt(5)/3 and 4asqrt(2)/9. The integrand is simply the sum of the volumes of the discs between the bases of the two domes. Integration gives us (81(pi)/16)(y-y^3/(3a^2)) between the limits, which I calculated to be 53.39 cu in approx. How many candies fit into this volume? We know the length of the candies is 1/2", but we don't know any other dimensions, so we don't know the volume of each candy. Also, the alignment of candies will improve the number of candies fitting into the bowl, but if this cannot be arranged, we have to assume they will be randomly orientated. We could make the assumption that they are cone-shaped with height 0.5" and base diameter 0.25" (radius 0.125" or 1/8") giving them a volume of (1/3)(pi)0.125^2*0.5=0.0082 cu in each approximately. So divide this into 53.39 and we get about 6,526, with very little space between the candies. A more realistic figure can be obtained by finding out how many candies fill a cubic inch when tightly packed. Two will fit lengthwise into an inch. Think of the candies as cuboids 1/4" square and 1/2" long (volume=1/4*1/4*1/2=1/32 cu in). That gives you 32 in a cubic inch, over 1,700 in a bowl. The cuboid is a sort of container that will hold just one candy and allow it some lengthwise freedom of movement. At the other extreme think of the candies as 1/2" cubes (volume=1/8 cu in) then only 8 will fit into a cubic inch and only about 427 will fit into the bowl. But the candies have greater freedom of movement in a cubic container. Let's say you can get 100 into one cubic inch packed tightly, then you would get about 5,340 in the bowl. Get a small box and pack in as many candies as you can. Measure the volume of the box (length*width*height). If N is the number of candies, then each has an average effective volume of (box volume)/N. Use this number to divide into the volume of the bowl.  
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a wire of length l is cut into two parts. One part is bent into a circle and the other into square

If the radius of the circle is d then its area is (pi)d^2 and its circumference is 2(pi)d, the length of wire we need to make the circle. The length of the remainder of the wire is l-2(pi)d, out of which we make the square. So the side of the square is a quarter of this perimeter, 1/4(l-2(pi)d), and the area of the square is the square of this side, 1/16(l-2(pi)d)^2. The sum of the areas of the circle and square is (pi)d^2+1/16(l-2(pi)d)^2. We need the minimum value of this expression, where the variable is d. So we differentiate it with respect to d. That's the same as getting the difference quotient. The expansion of 1/16(l-2(pi)d)^2 is l^2/16-1/4(pi)ld+1/4(pi)^2d^2. [Please distinguish between 1 and l in the following.] The difference quotient is zero at a maximum or minimum, so we have 2(pi)d-l/4(pi)+1/2(pi)^2d=0. We can take out (pi), leaving 2d-l/4+1/2(pi)d=0. Multiply through by 4 to get rid of the fractions: 8d-l+2(pi)d=0, from which d=l/(8+2(pi)). Half the length of the side of the square is 1/8(l-2(pi)d). If we substitute for d in this expression we get 1/8(l-2(pi)l/(8+2(pi)))  = l/8((8+2(pi)-2(pi))/(8+2(pi)) = l/8(8/(8+2(pi)) = l/(8+2(pi)) = d (QED). Therefore the radius of the circle = half the length of the side of the square is either a maximum or minimum value of the expression for the sum of the areas of the circle and square. We can see that this expression gets bigger as d gets bigger, because (pi)d^2 has a positive value always, so we do indeed have a minimum rather than a maximum. We can substitute d=l/(2(4+(pi))) in the expression for the sum of the areas and we get the minimum: (pi)d^2+1/16(l-2(pi)d)^2 = (pi)l^2/4(4+(pi))^2 + 1/16(l-2(pi)l/(2(4+(pi)))^2 = (pi)l^2/4(4+(pi))^2 + l^2/16(1-(pi)/(4+(pi))^2 = (pi)l^2/4(4+(pi))^2 + l^2/16(4+(pi)-(pi))^2(4+(pi))^2 = (pi)l^2/4(4+(pi))^2 + l^2/(4+(pi))^2 = l^2/(4(4+(pi))) or (l^2/4)*1/(4+(pi)) Sorry, it was getting difficult to represent the expressions using this tablet so I've had to accelerate the last bit! I hope I didn't make any mistakes!
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With using the following #'s 56,27,04,17,93 How do I find my answer of 41?

Let A, B, C, D, E represent the five numbers and OP1 to OP4 represent 4 binary operations: add, subtract, multiply, divide. Although the letters represent the given numbers, we don't know which unique number each represents. Then we can write OP4(OP3(OP2(OP1(A,B),C),D),E)=41 where OPn(x,y) represents the binary operation OPn between two operands x and y. We can also write: OP3(OP1(A,B),OP2(C,D))=OP4(41,E) as an alternative equation.  We are looking for a solution to either equation where 04...93 can be substituted for the algebraic letters and the four operators can be substituted for OPn. Because the operation is binary, requiring two operands, and there is an odd number of operands, OP4(41,E) must apply in either equation. So we only have to work out whether to use OP3(OP2(OP1(A,B),C),D)=OP4(41,E) or OP3(OP1(A,B),OP2(C,D))=OP4(41,E). Also we know that add and subtract have equal priority and multiply and divide have equal priority but a higher priority than add and subtract. 41 ia a prime number so we know that OP4 excludes division. If OP4 is subtraction, we will work with the absolute difference |41-E| rather than the actual difference. For addition and multiplication the order of the operands doesn't matter. This gives us a table (OP4 TABLE) of all possibilities,    04 17 27 56 93 + 45 58 68 97 134 - 37 24 14 11 52 * 164 697 1107 2296 3813 This table shows all possible results for the expression on the left-hand side of either of the two equations. The table below shows OPn(x,y): In this table the cells contain (R+C,|R-C|,RC,R/C or C/R) where R=row header and C=column header. The X cells are redundant. Now consider first: OP3(OP1(A,B),OP2(C,D)). To work out all possible results we have to take each value in each cell and apply operations between it and each value in all other cells not having the same row or column. For example, if we take 21 out of row headed 17, we are using (R,C)=(17,4), so we are restricted to operations involving (56,27), (93,27) and (93,56). We can only use the numbers in these cells, but the improper fractions can be inverted if necessary. If the result of the operation matches a number in the only cell in the OP4 TABLE with the remaining R and C values then we have solved the problem. As it happens, the sum of 21 from (17,4) and 37 from (93,56) is 58 which is in (+,17) of the OP4 TABLE. Unfortunately, the column number 17 is the same as the row number in (17,4). To qualify as a solution it would have to be in the 27 column of the OP4 TABLE. But let's see if the arithmetic is correct: (17+4)+(93-56)=21+37=58=41+17; so 41=(17+4)+(93-56)-17. Yes the arithmetic is correct, but 17 has been used twice and we haven't used 27. So the arithmetic actually simplifies to 41=4+93-56, because 17 cancels out and, in fact, we only used 3 numbers out of the 5. Still, you get the idea. (4*17)+(56-27)=68+29=97=41+56; 41=4*17+56-27-56 is another failed example because 93 is missing and 56 is duplicated, so this simplifies to 41=4*17-27, thereby omitting 56 and 93. Similarly, (27+4)+(93-27)=31+66=97=41+56; so 41=(27+4)+(93-27)-56=4+93-56, omitting 17 and 27. The question doesn't make it clear whether all 5 numbers have to be used just once to produce 41, or whether 41 has to be made up of only the given numbers, but not necessarily all of them. In the latter case, it's clear we have found some solutions. The method I used was to create a table made up of all possible OPn(x,y) as the row and column headers. Then I eliminated all cells where the (x,y) components of the row and column contained an element in common. Each uneliminated cell contained (sum,difference,product,quotient) values. When this table had been completed, I looked for occurrences of the numbers in the cells of the OP4 TABLE, and I highlighted them. The final task was to see if there were any highlighted cells such that the numbers that matched cells in the OP4 TABLE were in a column, the header of which made up the set of five given numbers. It happened that there were none, so OP3(OP1(A,B),OP2(C,D))=OP4(41,E) could not be satisfied. No arrangement of A, B, C, D or E with OP1 to OP4 could be found. More to follow...
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Which statements are true? Select all that apply.

True True False True False
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