Guide :

Isolate seven circles in three squares

Here is an isolation problem with an unusual requirement. There are seven circles which you must isolate in seven squares. However, you may draw only three squares, and no more than four squares may be the same size.

Research, Knowledge and Information :


Square the Circle - Home


Squaring the circle with ruler and ... He asserted that even God Himself could not square the circle using ... we need to isolate the following three shapes ...
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Performance Assessment Task Circle and Squares Grade 10 ...


the radius of a circle is perpendicular to the tangent where the radius intersects the circle. G‐C.3 Construct the inscribed and ... Circle and Squares
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Circles and Squares - DC Symbols


From the top we see three 20x20 squares; ... Note especially the three circles that define the ... To determine where these points fall we divide 1 by seven then ...
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BY COMPLETING THE SQUARE Circle Important Properties


Steps to verify the equation of a circle by completing the square: 1. Isolate the constant on one side of the ... Verifying circles by completing the square, page 3
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geometry - Three squares and three circles packed into a ...


If we have three squares and three circles inside a right triangle if the radius of the small circle is $19$ and the ... Optimal Packing of Seven Circles in a ...
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How many 3.75 inch circles will fit inside a 17 inch square ...


3.75″ diameter circle; square with 17″ sides; Shapes cannot be cut or deformed; Update Cancel. Answer Wiki. ... How many 1 inch circles will fit inside a 16 inch ...
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How to Solve Radical Equations: 12 Steps (with Pictures ...


How to Solve Radical Equations. ... The root is typically a square root, ... Isolate the other square root.
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4 Ways to Calculate the Area of a Circle - wikiHow


Nov 10, 2016 · A common problem in geometry class is to have you calculate the area of a circle based ... to isolate the variable . by ... The area of a circle is ...
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Suggested Questions And Answer :


Isolate seven circles in three squares

The picture shows one solution. The squares don't have to be tilted: However, this gives the appearance of more than four squares the same size.
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calculating probability and number of sets

Combinations give no relevance to order, but permutations do. If by set is meant those cards of the same colour then we can only form a set of three if we have three cards at least in that colour. There are three blue cards (bearing numbers 1, 4 and 7), and three green (2, 5 and 8). So only two such sets are possible. That's the combination. If we introduce order, then there are 6 ways of arranging the numbers 1, 4 and 7, and 6 ways for 2, 5 and 8, making 12 in all. If any group of three cards is considered to be a set then we can have nCr (where n=10 and r=3), that is 120, sets. If the order is important then we use nPr, that is 720. The difference is a factor of 6, which is the number of different ways of arranging (ordering) 3 objects. To calculate nCr, evaluate (n-1)(n-2)...(n-r+1)/fact(r). The denominator fact(r) is usually written r! (r factorial or r shriek) and consists of multiplying all the integers from 1 to r. nPr is just the numerator without the denominator, so nPr=r!*nCr. r! is the number of ways of arranging r different objects.
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. A car dealership has 556 new cars on its lot.

There seems to be one number missing: the number of black cars with automatic transmission and a sun roof, so we'll call this number X. It's helpful to use a diagram (Venn diagram). Draw a large circle (ellipse or other enclosure) containing three interlocking circles. I'll use the word "circle" to mean any completely enclosed area. The three circles represent black cars (B), automatics (A), and cars fitted with a sunroof (S) respectively. The large circle represents all the cars (C), so those that are not black, not automatic and have no sunroof are represented by the interior of the large circle outside the other three circles. We know there are 133 black cars, and 116 of these are automatic, so the remaining 17 must be manual transmission (non-automatic). We also know that 10 black cars have a sunroof, so 123 don't have a sunroof. And we know that X black automatics have a sunroof.  Of the 401 automatics, 42 have sunroofs, so 359 automatics have no sunroof; and of the 57 cars fitted with a sunroof, 42 are automatics, so 15 are manual. In the Venn diagram the intersection of the three sets are represented by the areas enclosed by two or three interlocking circles. I need a symbol to express intersection, normally represented by an inverted U symbol, which I don't have on my tablet, so I'll use ^, which is normally used to represent "to the power of". A^B means black automatics because it's the set of all black cars with automatic transmission, intersection of the black set B with the automatic set A. On the diagram it's the area enclosed by circles A and B. A^B^S is the area enclosed by the three interlocking circles and represents all black automatics with a sunroof. In all, the overlapping circles produce four enclosed areas: A^B, A^S, B^S and A^B^S. We can give values to these: 116, 42, 10 and X, respectively. Remember, we weren't given a value for X. We also have A=401, B=133, S=57. We can represent "not" by underlining a set, so, for example, A represents non-automatic and B^A would mean black cars without automatic transmission (AT), so B^A=17 because 17 cars are black without AT. Similarly, B^S=123. So B^A^S=A^B^S=116-X. That is, a) there are 116-X black automatics without a sunroof. There are 556-133=423 cars that are not black; 556-401=155 cars that are not automatic; and 556-57=499 cars without a sunroof. b) The final part of the question is easier to work out by looking at the diagram. We need to work out the total number of cars within the interlocking circles. We can't simply add the numbers of cars in the three circles and subtract the sum from 556, because the circles interlock. Take A and B for example. They interlock enclosing 116 cars (black automatics), so we deduct 116 from A and add the result to B; or we deduct 116 from B and add the result to A: 285+133=418=17+401.  If you look at the Venn diagram you'll see there are seven areas produced by the interlocking circles. Let's call the areas a, b, c, d, e, f and g, such that all automatics are given by a+d+g+e=401; black cars by b+d+g+f=133; and sunroofs by c+e+f+g=57. Diagrammatically, a is the set of automatics that are not black and have no sunroof; b the set of black cars that are neither automatic nor have a sunroof; and c the set of sunroofed cars that are neither black or automatic. There is an eighth area, h, outside of all the interlocking circles representing the residual cars that are not black, not automatic, and have no sunroof. Area g contains X cars (black automatics with a sunroof), so g=X; d+g=116; f+g=10; e+g=42; a+b+c+d+e+f+g+h=556. So d=116-X; f=10-X; e=42-X; a=401-(116+42-X)=43+X; b=133-(116+10-X)=7+X; c=57-(42-X+10)=5+X; (43+X)+(7+X)+(5+X)+(116-X)+(42-X)+(10-X)+X+h=556; so h=556-(223+X)=333-X. This is the number of cars that are not black, not automatic and have no sunroof. Although we don't know the value of X, we know it must be less than 10 because area f contains a number of cars that can't, obviously, be negative. The wording of the question suggests that X is greater than 1, so the possible answers range from 324 to 331 (1 Read More: ...

Give three sets A,B,C such that A∩B≠..... (elementary algebra)

The 3 sets A, B and C can be represented by 3 circles. The intersection of A and B is where circles A and B overlap. Similarly, B and C intersect (overlap). A, B and C overlap, so it would appear that A also overlaps C. We have three interlocking circles A, B and C. alpha can also be represented by a circle in isolation from the other three, no overlaps. So the Venn diagram contains 4 circles only three of which overlap (A, B and C). U is the universal set.
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How to find equation of the circle passing through (9,1), (-3,-1) and (4,5)

The equation of a circle is more specific: (x-h)^2+(x-k)^2=a^2, where (h,k) is the centre and a the radius. Plug in the points: (9-h)^2+(1-k)^2=(-3-h)^2+(-1-k)^2=(4-h)^2+(5-k)^2=a^2 This can be written (9-h)^2+(1-k)^2=(3+h)^2+(1+k)^2=(4-h)^2+(5-k)^2=a^2 Leaving a out of it for the moment we can use pairs of equations, using difference of squares: 12(6-2h)+2(-2k)=0, 36-12h-2k=0, 18-6h-k=0, so k=18-6h. 7(-1-2h)+6(-4-2k)=0, -7-14h-24-12k=0, -31-14h-12k=0 or 31+14h+12k=0, 31+14h+12(18-6h)=0, 31+14h+216-72h=0. 247-58h=0 so h=247/58 This looks suspiciously ungainly. So rather than continuing, I'm going to call the three points: (Q,R), (S,T), (U,V) to provide a general answer to all questions of this sort. (Q-h)^2+(R-k)^2=(S-h)^2+(T-k)^2=(U-h)^2+(V-k)^2=a^2 [(equation 1)=(equation 2)=(equation 3)=a^2] Take the equations in pairs and temporarily ignore a^2. Equations 1 and 2: (Q-S)(Q+S-2h)+(R-T)(R+T-2k)=0 Q^2-S^2-2h(Q-S)+R^2-T^2-2k(R-T)=0 2k(R-T)=Q^2+R^2-(S^2+T^2)-2h(Q-S), so k=(Q^2+R^2-(S^2+T^2)-2h(Q-S))/(2(R-T)) Equations 2 and 3: S^2-U^2-2h(S-U)+T^2-V^2-2k(T-V)=0, so k=(S^2+T^2-(U^2+V^2)-2h(S-U))/(2(T-V)) At this point, we can substitute for k and end up with an equation involving the unknown h only: (Q^2+R^2-(S^2+T^2)-2h(Q-S))/(2(R-T))=(S^2+T^2-(U^2+V^2)-2h(S-U))/(2(T-V)) (T-V)(Q^2+R^2-(S^2+T^2)-2h(Q-S))=(R-T)(S^2+T^2-(U^2+V^2)-2h(S-U)) (T-V)(Q^2+R^2-(S^2+T^2))-2h(T-V)(Q-S)=(R-T)(S^2+T^2-(U^2+V^2))-2h(R-T)(S-U) 2h((R-T)(S-U)-(T-V)(Q-S))=(R-T)(S^2+T^2-(U^2+V^2))-(T-V)(Q^2+R^2-(S^2+T^2)) h=((R-T)(S^2+T^2-(U^2+V^2))-(T-V)(Q^2+R^2-(S^2+T^2)))/((R-T)(S-U)-(T-V)(Q-S)). Once h is found we can calculate k, and then we can substitute the values for h and k in any equation to find a^2 which is equal to each of the three equations. When we use Q=9, R=1, S=-3, T=-1, U=4, V=5 or -5, we appear to get very ungainly solutions. One way to find which values need to be changed may be to plot the values and work out where the circle should fit and where its centre has less complex values. If V=6 or -6, there is a simple solution: (x-3)^2+y^2=37 (centre at (3,0), a^2=37=6^2+1^2. (-3-3)^2=(9-3)^2=6^2; (-6)^2=6^2; (2-3)^2=(4-3)^2=1; (-1)^2=1^2 shows the combination of points that would lie on the circle: (9,1), (9,-1), (-3,1), (-3,-1), (4,6), (4,-6), (2,6), (2,-6) and this includes points A and B, but not C. Note that the equation k=18-6h is valid for h=3, k=0.
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Find the centre of a circle that passes through the points D (-5,4), E (-3,8), and F (1,6).

A circle with centre (h,k) and radius a has the equation (x-h)^2+(y-k)^2=a^2. If we substitute x and y for the three points, we have three equations: a) (-5,4) (5+h)^2+(4-k)^2=a^2 [(-5-h)^2=(5+h)^2 because -1^2=+1^2=1] b) (-3,8) (3+h)^2+(8-k)^2=a^2=(5+h)^2+(4-k)^2 (3+h)^2-(5+h)^2+(8-k)^2-(4-k)^2=0 We have difference of squares twice on the left, so we can make use of A^2-B^2=(A-B)(A+B): -2(8+2h)+4(12-2k)=0, -16-4h+48-8k=0, 32=4h+8k, 8=h+2k, so k=(8-h)/2. c) (1,6) (1-h)^2+(6-k)^2=a^2=(5+h)^2+(4-k)^2=a^2 6(-4-2h)+2(10-2k)=0, -24-12h+20-4k=0, -4-12h-4k=0, -1-3h-k=0=-1-3h-(8-h)/2, so 2+6h+8-h=0 and h=-2. So k=5. Therefore, a^2=3^2+1=10. The equation of the circle becomes (x+2)^2+(y-5)^2=10.    
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how do you do a SA cyliander?

how do you do a SA cyliander? remember there are three thing to account for. the top and bottom of the cylinder (area of a circle).  A = pi * radius^2 (you need 2 of these). the side of the cylinder is a square or rectangle (circumference of the circle * height) circumference is 2 * pi * radius
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what are the possible combination of making 30 with 1, 3, 5, 7, 9

I started off by figuring how many ways to use nines: 30   3 nines, 27 total   2 nines, 18 total   1 nine, 9 total   0 nines, 0 total I then figured how many ways to use sevens: 30   3 nines, 27 total     0 sevens, 27 total   2 nines, 18 total     1 seven, 25 total     0 sevens, 18 total   1 nine, 9 total     3 sevens, 30 total     2 sevens, 23 total     1 seven, 16 total     0 sevens, 9 total   0 nines, 0 total     4 sevens, 28 total     3 sevens, 21 total     2 sevens, 14 total     1 seven, 7 total     0 sevens, 0 total I then figured how many ways to use fives, then threes, ending up with this: 30   3 nines, 27 total     0 sevens, 27 total       0 fives, 27 total         1 three, 30 total         0 threes, 27 total   2 nines, 18 total     1 seven, 25 total       1 five, 30 total         0 threes, 30 total       0 fives, 25 total         1 three, 28 total         0 threes, 25 total     0 sevens, 18 total       2 fives, 28 total         0 threes, 28 total       1 five, 23 total         2 threes, 29 total         1 three, 26 total         0 threes, 23 total       0 fives, 18 total         4 threes, 30 total         3 threes, 27 total         2 threes, 24 total         1 three, 21 total         0 three, 18 total   1 nine, 9 total ​    3 sevens, 30 total       0 fives, 30 total         0 threes, 30 total     2 sevens, 23 total       1 five, 28 total         0 threes, 28 total       0 fives, 23 total         2 threes, 29 total         1 three, 26 total         0 threes, 23 total     1 seven, 16 total       2 fives, 26 total         1 three, 29 total         0 threes, 26 total       1 five, 21 total         3 threes, 30 total         2 threes, 27 total         1 three, 24 total         0 threes, 21 total       0 fives, 16 total         4 threes, 28 total         3 threes, 25 total         2 threes, 22 total         1 three, 19 total     0 threes, 16 total     0 sevens, 9 total       4 fives, 29 total     0 threes, 29 total       3 fives, 24 total     2 threes, 30 total     1 three, 27 total     0 threes, 24 total       2 fives, 19 total     3 threes, 28 total     2 threes, 25 total     1 three, 22 total     0 threes, 19 total       1 five, 14 total     5 threes, 29 total     4 threes, 26 total         3 threes, 23 total     2 threes, 20 total     1 three, 17 total     0 threes, 14 total       0 fives, 9 total     7 threes, 30 total     6 threes, 27 total     5 threes, 24 total     4 threes, 21 total     3 threes, 18 total     2 threes, 15 total     1 three, 12 total     0 threes, 9 total   0 nines, 0 total     4 sevens, 28 total       0 fives, 28 total     0 threes, 28 total     3 sevens, 21 total       1 five, 26 total     1 three, 29 total     0 threes, 26 total       0 fives, 21 total     3 threes, 30 total     2 threes, 27 total     1 three, 24 total     0 threes, 21 total     2 sevens, 14 total       3 fives, 29 total     0 threes, 29 total       2 fives, 24 total     2 threes, 30 total     1 three, 27 total     0 threes, 24 total       1 five, 19 total     3 threes, 28 total     2 threes, 25 total     1 three, 22 total     0 threes, 19 total       0 fives, 14 total     5 threes, 29 total     4 threes, 26 total     3 threes, 23 total     2 threes, 20 total     1 three, 17 total     0 threes, 14 total     1 seven, 7 total       4 fives, 27 total     1 three, 30 total     0 threes, 27 total       3 fives, 22 total     2 threes, 28 total     1 three, 25 total     0 threes, 22 total       2 fives, 17 total     4 threes, 29 total     3 threes, 26 total     2 threes, 23 total     1 three, 20 total     0 threes, 17 total       1 five, 12 total     6 threes, 30 total     5 threes, 27 total     4 threes, 24 total     3 threes, 21 total     2 threes, 18 total     1 three, 15 total     0 threes, 12 total       0 fives, 7 total     7 threes, 28 total     6 threes, 25 total     5 threes, 22 total     4 threes, 19 total     3 threes, 16 total     2 threes, 13 total     1 three, 10 total     0 threes, 7 total     0 sevens, 0 total       6 fives, 30 total     0 threes, 30 total       5 fives, 25 total     1 three, 28 total     0 threes, 25 total       4 fives, 20 total     3 threes, 29 total     2 threes, 26 total     1 three, 23 total     0 threes, 20 total       3 fives, 15 total     5 threes, 30 total     4 threes, 27 total     3 threes, 24 total     2 threes, 21 total     1 three, 18 total     0 threes, 15 total       2 fives, 10 total     6 threes, 28 total     5 threes, 25 total     4 threes, 22 total     3 threes, 19 total     2 threes, 16 total     1 three, 13 total     0 threes, 10 total       1 five, 5 total     8 threes, 29 total     7 threes, 26 total     6 threes, 23 total     5 threes, 20 total     4 threes, 17 total     3 threes, 14 total     2 threes, 11 total     1 three, 8 total     0 threes, 5 total       0 fives, 0 total     10 threes, 30 total     9 threes, 27 total     8 threes, 24 total     7 threes, 21 total     6 threes, 18 total     5 threes, 15 total     4 threes, 12 total     3 threes, 9 total     2 threes, 6 total     1 three, 3 total     0 threes, 0 total (more to follow)
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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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Creata a fraction story

FIVE LITTLE PIGS GO TO THE MALL Mama Pig gave her five little pigs seven and a half dollars between them to spend at the mall. It was a cold day, twenty-three Fahrenheit, minus five Celsius, or five degrees below freezing. Off they trotted at a quarter to three in the afternoon. "How far is it?" the youngest pig asked after a while. "One point seven five miles from home," said the eldest. "What does that mean?" asked the youngest. "Well," explained the eldest, "if we divide the distance into quarter miles, it's seven quarters." "How long will it take to get there?" asked the pig in the middle. Her twin sister replied, "It's five past three now, so that means we've taken twenty minutes to get here. Remember the milestone outside our house? There's another one here, so we've come just one mile and we've taken a third of an hour. [That means our speed must be three miles an hour.] "How much longer?" the youngest asked. "Three quarters of a mile to go," the next youngest started. "Yes," said the eldest, "we get the time by dividing distance by speed, so that means three quarters divided by three, which is one quarter of an hour [which is fifteen minutes]." ["So what time will we arrive?" the youngest asked. "About twenty past three," all the other pigs replied together.] "That's thirty-five minutes altogether," the eldest continued, "which means that - let me see - seven quarters divided by three is seven twelfths of an hour. One twelfth of an hour is five minutes [so seven twelfths is thirty-five minutes]. Yes, that's right." When they got to the mall, it had started to snow. Outside there was a big thermometer and a sign: "COME ON IN. IT'S WARMER INSIDE!" The eldest observed: "It shows temperature in Fahrenheit and Celsius. See, there's a scale on each side of the gauge. It's warmer now than it was when we left home. The scales are divided into tenths of a degree. It says twenty-seven Fahrenheit exactly, and, look, that's the same as minus two point eight Celsius," speaking directly to the youngest, "because the top of the liquid is about eight divisions between minus two and minus three. Our outside thermometer at home is digital, but this is analogue." The youngest stuttered: "What's 'digital' and 'analogue'?" "Well," began one of the twins, "your watch is analogue, because it has fingers that move round the watch face. Our thermometer at home is digital, because it just shows the temperature in numbers." ["Yes," said the eldest, "it would show thirty-seven point zero degrees Fahrenheit, four degrees warmer, and minus two point eight Celsius, two point two degrees warmer than when we left."] Continued in comment... 
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