I hope you find it helpful
The solution for this problem is:
If there is 60 platters of B at a cost of $720:
(220 - 60 x 3) / 4 = 10 platters of A to make up for the deficit in hamburgers
(270 - 60 x 4) / 3 = 10 platters of A to make up for the deficit in hot dogs
(250 - 60 x 5) / 2 = 0 platters of A since there is no deficit in pigs feet
So 10 platters A are required at a cost of $150. $720 + $150 = for a total minimum cost of $870.
Answer:
CD = √11 and CE = √11
Step-by-step explanation:
We know that m∠D is 45° (by using the sum of interior angles in a triangle) so therefore, ΔDCE is a 45 - 45 - 90 triangle (the 45, 45, and 90 refer to the angle measures). The ratio of sides in a 45 - 45 - 90 triangle is 1 : 1 : √2 where the 1s are the sides and the √2 is the hypotenuse. We need to solve for x in x : x : √22. If you notice that √22 = √2 * √11, we can use this to find x, therefore, x = 1 * √11 = √11 so CD = √11 and CE = √11.
![f(x)=(x+4)(x-2)(x+4)(x+4)\\f(x)=(x+4)^3(x-2)=0\\(x+4)^3=0~|~(x-2)=0\\(x+4)(x+4)(x+4)=0~|~x=2\\x=-4,2](https://tex.z-dn.net/?f=f%28x%29%3D%28x%2B4%29%28x-2%29%28x%2B4%29%28x%2B4%29%5C%5Cf%28x%29%3D%28x%2B4%29%5E3%28x-2%29%3D0%5C%5C%28x%2B4%29%5E3%3D0~%7C~%28x-2%29%3D0%5C%5C%28x%2B4%29%28x%2B4%29%28x%2B4%29%3D0~%7C~x%3D2%5C%5Cx%3D-4%2C2)
The multiple zero is x = -4.
The multiplicity is 2 because there are two values of x.
<u>Given</u><u> </u><u>Information</u><u> </u><u>:</u><u>-</u>
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- A polygon with 10 sides ( Decagon )
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<u>To</u><u> </u><u>Find</u><u> </u><u>:</u><u>-</u>
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- The value of one of the exterior angles
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<u>Formula</u><u> </u><u>Used</u><u> </u><u>:</u><u>-</u>
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![\qquad \diamond \: \underline{ \boxed{ \pink{ \sf Exterior ~angle = \dfrac {360^\circ}{no. ~of~sides}}}} \: \star](https://tex.z-dn.net/?f=%20%5Cqquad%20%5Cdiamond%20%5C%3A%20%20%5Cunderline%7B%20%5Cboxed%7B%20%5Cpink%7B%20%5Csf%20Exterior%20~angle%20%3D%20%5Cdfrac%20%7B360%5E%5Ccirc%7D%7Bno.%20~of~sides%7D%7D%7D%7D%20%5C%3A%20%20%5Cstar)
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<u>Solution</u><u> </u><u>:</u><u>-</u>
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Putting the given values, we get,
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![\sf \dashrightarrow Exterior ~angle = \dfrac{360 ^\circ}{10} \: \: \\ \\ \\ \sf \dashrightarrow Exterior ~angle = \frac{36 \cancel{0}^\circ}{ \cancel{10}} \: \: \\ \\ \\ \sf \dashrightarrow Exterior ~angle = \underline{ \boxed{ \frak{ \red{36^\circ}}}} \: \star \\ \\](https://tex.z-dn.net/?f=%5Csf%20%5Cdashrightarrow%20Exterior%20~angle%20%3D%20%20%5Cdfrac%7B360%20%20%5E%5Ccirc%7D%7B10%7D%20%5C%3A%20%20%5C%3A%20%20%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20%5Csf%20%5Cdashrightarrow%20Exterior%20~angle%20%3D%20%20%5Cfrac%7B36%20%5Ccancel%7B0%7D%5E%5Ccirc%7D%7B%20%5Ccancel%7B10%7D%7D%20%5C%3A%20%20%5C%3A%20%20%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20%5Csf%20%5Cdashrightarrow%20Exterior%20~angle%20%3D%20%20%5Cunderline%7B%20%5Cboxed%7B%20%5Cfrak%7B%20%5Cred%7B36%5E%5Ccirc%7D%7D%7D%7D%20%5C%3A%20%5Cstar%20%5C%5C%20%20%5C%5C%20)
Thus, the value of the exterior angles of a Decagon is 36°.
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![\underline{ \rule{227pt}{2pt}} \\ \\](https://tex.z-dn.net/?f=%20%5Cunderline%7B%20%5Crule%7B227pt%7D%7B2pt%7D%7D%20%5C%5C%20%20%5C%5C%20)