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madam [21]
1 year ago
12

Find the area of the isosceles triangle.

Mathematics
1 answer:
cricket20 [7]1 year ago
5 0

Check the picture below.

well, we know the triangle is an isosceles, so it has twin sides coming from the "vertex" down to the "base", running an angle bisector from the "vertex" will give us a perpendicular to the "base", let's find its height.

\textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2 - a^2}=b \qquad \begin{cases} c=\stackrel{hypotenuse}{13}\\ a=\stackrel{adjacent}{5}\\ b=\stackrel{opposite}{h}\\ \end{cases} \\\\\\ \sqrt{13^2 - 5^2}=h\implies 12=h

so we simply need to get the area of a triangle whose base is 10 and height is 12.

A=\cfrac{1}{2}(\underset{b}{10})(\underset{h}{12})\implies \boxed{A=60}

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Step-by-step explanation:

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What is the answer to this problem<br> 2(6y-2)-3y=2
tester [92]

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5 0
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Assume lim f(x)-6 and lim g(x)-9. Compute the following limit and state the limit laws used to justify the computation.
emmasim [6.3K]

Answer:

4

Step-by-step explanation:

<h3><u>some relevant limit laws</u></h3>

lim C = C where c is a constant.

lim( f(x)  + g(x)) =lim f(x) + lim g(x)

lim( f(x)g(x)) =lim f(x) * lim g(x)

lim( cg(x)) =clim g(x)

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lim( f(x))^2 = (lim f(x) )^2

lim  square root( f(x)) = square root(lim f(x) )

\lim_{n \to 3} g(x)  = 9\\\\\lim_{n \to 3} f(x)  = 6\\\\ \lim_{n \to 3} \sqrt[3]{f(x)g(x) + 10} \\\\ = \lim_{n \to 3} \sqrt[3]{f(x)g(x) + 10}\\\\= \sqrt[3]{lim_{n \to 3}f(x) \times lim_{n \to 3}g(x) + 10}\\\\= \sqrt[3]{6 \times 9 + 10}\\\\= \sqrt[3]{64}

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4 0
3 years ago
a 18 ft tall statue standing next to a globe casts a 12 ft shadow. Of the globe casts a shadow that is 2 ft ling, then how tall
Igoryamba
<h3>Answer:</h3>

3 ft

<h3>Step-by-step explanation:</h3>

The statue's height is 1.5 times the length of its shadow, so we expect the same relationship for the globe.

... 1.5 × 2 ft = 3 ft

_____

<em>Comment on the problem</em>

As a practical matter, with the sun high enough in the sky to cast a shadow shorter than the object's height, it will be quite difficult to measure the length of the shadow of the point at the top of the globe. The shadow of other parts of the globe will interfere.

4 0
3 years ago
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