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3 years ago

9

Mrs. Freeman is trying to determine if her new desk will fit through her classroom door. Her desk of her is 59 inches wide. She

knows that the height of the door is 80 inches and the diagonal measure of the door is 100 inches. Determine the width of her door de ella and whether or not her new desk will fit. Justify your conclusion.

1 answer:

iren [92.7K]3 years ago

8 0

Answer: x = 60

Step-by-step explanation:

the equation is a^2 + b^2 = c^2.

If you draw out the problem, you plug in the information into the formula, so it would be...

x^2 + 80^2 = 100^2.

80^2 = 6,400 and 100^2 = 10,000.

x^2 + 6,400 = 10,000.

You subtract 6,400 from itself and 10,000 and then you get...

x^2 = 3,600.

You square both of those and your final answer will be x = 60.

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Evaluate if m = 16 and p = 2. 1/2m + 3p A) 10 B) 14 C) 18 D) 48

Zepler [3.9K]

Answer:

B) 14

Step-by-step explanation:

We can plug in the values of m and p into the equation:

1/2 m + 3p = 1/2 (16) + 3 (2) = 8 + 6 = 14.

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3 years ago

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

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

We have

$\sqrt{x^2-x}=\sqrt{x^2}\sqrt{1-\dfrac1x}$
$\sqrt{x^2-x}=|x|\sqrt{1-\dfrac1x}$
$\sqrt{x^2-x}=x\sqrt{1-\dfrac1x}$

since $x\to\infty$ forces us to have $x>0$ . And as $x\to\infty$ , the $\dfrac1x$ term is "negligible", so really $\sqrt{x^2-x}\approx x$ . We can then treat the limand like

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

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.

$\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-4x-b)=0$

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

We write

$(\sqrt{x^2-x}-x)\cdot\dfrac{\sqrt{x^2-x}+x}{\sqrt{x^2-x}+x}=\dfrac{(x^2-x)-x^2}{\sqrt{x^2-x}+x}=-\dfrac x{x\sqrt{1-\frac1x}+x}=-\dfrac1{\sqrt{1-\frac1x}+1}$

Now as $x\to\infty$ , we see this expression approaching $-\dfrac12$ , so that

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

So one asymptote of the hyperbola is the line $y=4x-2$ .

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 $b=-1$ .

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

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

$(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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This time the limit is $\dfrac12$ , so

$\dfrac12=\dfrac{b+1}2\implies b=0$

which means the other asymptote is the line $y=0$ .

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Taya2010 [7]

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3 years ago

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Answer:

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Step-by-step explanation:

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In this case, the width is $3y+1$ miles and the length is $23$ miles. So, the area is $23(3y+1)=23*3*y+23*1=69*y+23$ .

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stiks02 [169]

Answer:

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Step-by-step explanation:

That's my guess.

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