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never [62]
3 years ago
15

Find complement and supplement of the angle. 75 degrees

Mathematics
1 answer:
kow [346]3 years ago
4 0
x-\ measure\ of\ supplement\ angle\\\\
180=75+x\ \ \ | subtract\ 75\\\\
105=x\\\\
y-\ measure \ of\ complement\ angle\\\\
90=y+75\ \ \ | subtract\ 75\\\\
y=15\\\\Complement\ angle\ is\ 15^{\circ}\ and\ supplement\ 105^{\circ}
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Ples help me find slant assemtotes
FrozenT [24]
A polynomial asymptote is a function p(x) such that

\displaystyle\lim_{x\to\pm\infty}(f(x)-p(x))=0

(y+1)^2=4xy\implies y(x)=2x-1\pm2\sqrt{x^2-x}

Since this equation defines a hyperbola, we expect the asymptotes to be lines of the form p(x)=ax+b.

Ignore the negative root (we don't need it). If y=2x-1+2\sqrt{x^2-x}, then we want to find constants a,b such that

\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-ax-b)=0

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2x-1+2x-ax-b=(4-a)x-(b+1)

which tells us that we would choose a=4. You might be tempted to think b=-1, but that won't be right, and that has to do with how we wrote off the "negligible" term. To find the actual value of b, we have to solve for it in the following limit.

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-\dfrac12=\dfrac{b+1}2\implies b=-2

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The other asymptote is obtained similarly by examining the limit as x\to-\infty.

\displaystyle\lim_{x\to-\infty}(2x-1+2\sqrt{x^2-x}-ax-b)=0

\displaystyle\lim_{x\to-\infty}(2x-2x\sqrt{1-\frac1x}-ax-(b+1))=0

Reduce the "negligible" term to get

\displaystyle\lim_{x\to-\infty}(-ax-(b+1))=0

Now we take a=0, and again we're careful to not pick b=-1.

\displaystyle\lim_{x\to-\infty}(2x-1+2\sqrt{x^2-x}-b)=0

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(x+\sqrt{x^2-x})\cdot\dfrac{x-\sqrt{x^2-x}}{x-\sqrt{x^2-x}}=\dfrac{x^2-(x^2-x)}{x-\sqrt{x^2-x}}=\dfrac
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Answer:

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This formula is :

V=h*w*l

Lets plug in what we know, and solve for the missing variable.

We know that height(h) is 7.

We know that width(w) is 9.

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So:

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To find length(l), we must iscolate it.

To do this, we must get rid of the 7 and 9.

So lets start by dividing 378 by 9:

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Answer:

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