16. Solve the equation. Check the solution.

Mathematics

1 answer:

harina [27]3 years ago

8 0

Hi there

a/(a² - 36) + 2/(a-6) = 1/(a+6)
a/(a+6)(a-6) + 2/(a-6) = 1/(a+6)
Now we need to multiply the terms by (a+6)(a-6)
a + 2(a+6) = 1(a-6)
a + 2a + 12 = a - 6
3a + 12 = a - 6
3a - a = -6 - 12
2a = -18
Divide both sides by 2
2a/2 = -18/2
a = -9
Thus, The correct option is A

I hope that's help !

Good luck !

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Find the missing length the triangles are similar.

ch4aika [34]

Answer:

Step-by-step explanation:

The missing length is 13 because,

Lets say the top triangle is A and the bottom triangle is B.

Triangle A gives us the side GF, and Triangle B gives us the sides TU and ST. Since the triangles are similar(as stated in the problem), we can pair 2 sides GF(A) and TU(B) which is 11:22.(one way I usually figure which sides are similar is by first- matching the hypotenuse, then checking which of the remaining two is longer.. if that made any sense). You can see that their relationship is x2 or /2 (In another word, from A to B is multiplication- ex: 11 * 2 is 22, and from B to A is division- ex 22/2 is 11.) Since the missing number is the hypotenuse of triangle A and you know the Hypotenuse of triangle B all you have to do is divide side TS by 2 to get side SF. So the missing side is 13.

Answer: 13

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3 years ago

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Stella [2.4K]

Answer:

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Step-by-step explanation:

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3 years ago

If <img src="https://tex.z-dn.net/?f=%20300cm%5E%7B2%7D%20" id="TexFormula1" title=" 300cm^{2} " alt=" 300cm^{2} " align="absmid

Artist 52 [7]

Check the picture below. Recall, is an open-top box, so, the top is not part of the surface area, of the 300 cm². Also, recall, the base is a square, thus, length = width = x.

$\bf \textit{volume of a rectangular prism}\\\\
V=lwh\quad 
\begin{cases}
l = length\\
w=width\\
h=height\\
-----\\
w=l=x
\end{cases}\implies V=xxh\implies \boxed{V=x^2h}\\\\
-------------------------------\\\\
\textit{surface area}\\\\
S=4xh+x^2\implies 300=4xh+x^2\implies \cfrac{300-x^2}{4x}=h
\\\\\\
\boxed{\cfrac{75}{x}-\cfrac{x}{4}=h}\\\\
-------------------------------\\\\
V=x^2\left( \cfrac{75}{x}-\cfrac{x}{4} \right)\implies V(x)=75x-\cfrac{1}{4}x^3$

so.. that'd be the V(x) for such box, now, where is the maximum point at?

$\bf V(x)=75x-\cfrac{1}{4}x^3\implies \cfrac{dV}{dx}=75-\cfrac{3}{4}x^2\implies 0=75-\cfrac{3}{4}x^2
\\\\\\
\cfrac{3}{4}x^2=75\implies 3x^2=300\implies x^2=\cfrac{300}{3}\implies x^2=100
\\\\\\
x=\pm10\impliedby \textit{is a length unit, so we can dismiss -10}\qquad \boxed{x=10}$

now, let's check if it's a maximum point at 10, by doing a first-derivative test on it. Check the second picture below.

so, the volume will then be at $\bf V(10)=75(10)-\cfrac{1}{4}(10)^3\implies V(10)=500 \ cm^3$