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Zanzabum
3 years ago
9

A man whose height is 2.1 m casts a shadow that is 3.2 m in length. What is the approximate height of a building that casts a sh

adow of 11.5 m?
a. 10.4 m
b. 9.6 m
c. 7.5 m
d. 17.5 m

Mathematics
1 answer:
Colt1911 [192]3 years ago
6 0

Answer:

ok so first

2.1 divided by 3.2 =0.65625

0.65625*11.5=7.546875

So the answer is B

Hope This Helps!!!

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tensa zangetsu [6.8K]

The two bottom graphs demonstrate translations.

<h3>Which figures demonstrate a translation?</h3>

We will have a translation only if:

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The images where the figures are only moved a little bit are the ones that demonstrate just a translation, and these are the two lower ones.

If you want to learn more about translations:

brainly.com/question/24850937

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8 0
2 years ago
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il63 [147K]

Answer:

x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}

Step-by-step explanation:

w=-5\left(x-8\right)\left(x+4\right)\\\mathrm{Expand\:}-5\left(x-8\right)\left(x+4\right):\quad -5x^2+20x+160\\w=-5x^2+20x+160\\Switch\:sides\\-5x^2+20x+160=w\\\mathrm{Subtract\:}w\mathrm{\:from\:both\:sides}\\-5x^2+20x+160-w=w-w\\Simplify\\-5x^2+20x+160-w=0\\Solve\:with\:the\:quadratic\:formula\\\mathrm{Quadratic\:Equation\:Formula:}\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=-5,\:b=20,\:c=160-w:\quad x_{1,\:2}=\frac{-20\pm \sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}\\x=\frac{-20+\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20+\sqrt{-20w+3600}}{10}\\x=\frac{-20-\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20-\sqrt{-20w+3600}}{10}\\The\:solutions\:to\:the\:quadratic\:equation\:are\\x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}

6 0
3 years ago
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