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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

## Research, Knowledge and Information :

### 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, ...

### 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 ...

### 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 ...

### 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?

### 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 ...

### 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 ...

### 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 ...

### 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 ...

### 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.

## 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

### 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

### 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

### 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

### 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.

### 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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