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ella [17]
3 years ago
8

John has 2456 pennies in his coin collection. He has the same number of pennies in each of 3 boxes. Estimate to the nearest hund

red the number of pennies in each box.
Mathematics
1 answer:
zhenek [66]3 years ago
6 0
2456 / 3 = 818
Rounded = 800

Answer 800 Pennies in each box.
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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

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
 x{x-(-x)\sqrt{1-\frac1x}}=\dfrac1{1+\sqrt{1-\frac1x}}

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.
4 0
3 years ago
Identify the coefficient in the term 7x^2y^3
Advocard [28]

Answer:

<h3>LARGEST COEFFICIENT = 7</h3><h3>SMALLEST COEFFICIENT=2</h3><h3>STANDARD FORM=7,0,0,0,3,2,0,0</h3><h3>COEFFICIENTS =7,3,2</h3>
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2 years ago
Round 3.751 to the nearest tenth.
ss7ja [257]

Answer: It's 3.8


Step-by-step explanation:


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3 years ago
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Correct answer gets brainliest
rosijanka [135]

Answer:

yea its 828

Step-by-step explanation:

Its alotta work that ion wanna type but that should b the answer

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3 years ago
Sam had some money in his pocket, and he found another $6.50 in his dresser drawer. He then had a total of $19.75. Let p represe
Bingel [31]
P=19.75−6.50
P=13.25
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