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pashok25 [27]
3 years ago
14

I need help asap!

Mathematics
2 answers:
777dan777 [17]3 years ago
7 0

Answer:

240

Step-by-step explanation:

Miss Jennifer is buying pencils in packs of 12, notebooks in packs of 16, and red pens in boxes of 20. She needs to have the same number of pencils, notebooks, and red pens. What is the least number of packs of each type should she buy?

So you know 12, 16, and 20 are even so you would find the LCM (Least Common Multiple) so 12 is 12, 24, 36, 48, 60, 72, 84, 96, 108, 120 so I know 20 and 12 go in 120 because 20, 40, 60, 80, 100, 120 but I know 16 is not but 12, 16, 20 is able to got toe 240 because 16 is able to and I know double of 120 is 240 and since 20 and 12 can go to 120 my answer is 240.

                               

balandron [24]3 years ago
4 0

Answer:

Miss Jennifer is buying pencils in packs of 12, notebooks in packs of 16, and red pens in boxes of 20. She needs to have the same number of pencils, notebooks, and red pens. What is the smallest number of packs of each type that she should buy?

Step-by-step explanation:

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Consider using the Law of Cosines, because the lengths of three sides are given and the largest angle is the one to be approximated. This angle will be opposite the longest side, that is, opposite the 420-foot side.

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\mathbf{\iiint_E  E \sqrt{x^2+y^2} \ dV =\dfrac{81 \  \pi}{80}}

Step-by-step explanation:

The Cylindrical coordinates are:

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From the question, on the xy-plane;

9 -16 (x^2 + y^2) = 0 \\ \\  16 (x^2 + y^2)  = 9 \\ \\  x^2+y^2 = \dfrac{9}{16}

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∴

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0}  \ \ \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2})  \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0}  ( 9r^2-16r^4})  \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0}   ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0}  \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0}   ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0}  \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0}   ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}

\iiint_E  E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi

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