2 years ago

Any one that knows this please help thanks!

Mathematics

1 answer:

nignag [31]2 years ago

3 0

You just need to plug the values: you have to multiply $x$ and $y^2$ , and then divide the result by -3.

So, if $y=2$ , we have $y^2=4$

Which means that $xy^2 = 5\cdot 4 = 20$

And the final answer is $\frac{20}{-3}$

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Two vertical poles of length 10 and 12 feet, respectively, stand 17 feet apart. A cable reaches from the top of one pole to the

Marat540 [252]

Answer:

$L=\sqrt{x^{2}+100}+\sqrt{x^{2}-34x+433} ft$

Step-by-step explanation:

let length of calble = L

distance from 10-foot pole = x ft

Moving from the top of the 10-foot pole to point x, we have

From the diagram that

$x^{2} +10^{2}=L_{1} ^{2} \\x^{2} + 100 = L_{1} ^{2}$

Take the square root of both sides

$\sqrt{x^{2}+100}=\sqrt{L_{1} ^{2}}\\ \sqrt{x^{2}+100}= L_{1}$

Moving from point x to the top of the 12-foot pole, we have

From the diagram that

$(17-x)^{2}+12^{2}=L_{2} ^{2} \\289-34x+x^{2} +144=L_{2} ^{2}\\x^{2} -34x+433=L_{2} ^{2}$

Take the square root of both sides

$\sqrt{x^{2}-34x+433}=\sqrt{L_{2} ^{2}} \\\sqrt{x^{2}-34x+433}=L_{2}$

Amount of cable used, L = L1 + L2

$L=\sqrt{x^{2}+100}+\sqrt{x^{2}-34x+433}$

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3 years ago

Write 1.33 as a percentage

ryzh [129]

1.33 as a percent would be ...................
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Which of the following fractions compares BC to BD? 2/3 5/2 2/5 3/2

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BC = 2 tiles

BD = 3 tiles

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