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ivanzaharov [21]
3 years ago
6

How do i solve this question

20%20-%2080t%20%2B%201200%29" id="TexFormula1" title="h = \frac{1}{16} ( {t}^{2} - 80t + 1200)" alt="h = \frac{1}{16} ( {t}^{2} - 80t + 1200)" align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
sattari [20]3 years ago
7 0

Answer:

h=1/16t^2-5t+75

Step-by-step explanation:

solve for h

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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
How much will a person pay for 7.5 pounds of bananas at a price of $2.07 per pound?
igomit [66]

Answer:

$15.57

Step-by-step explanation:

2.07 times 7.50 = $15.57!

7 0
3 years ago
Triangle DEF is similar to triangle PQR
almond37 [142]
It is a beacuse it’s is q that’s all
8 0
3 years ago
HELP ME PLEASEEE! Ty:)
Jlenok [28]
I can help you if you mark me as braì LG
5 0
3 years ago
What is the answer and work to this problem. <br> 4x+2y=14<br> 2x+y=7
Nimfa-mama [501]

The given system of equation has no solution.

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

<u><em>step 1</em></u><em> :</em>

The given equations are 4x + 2y = 14 and 2x + y = 7.

<u><em>step 2</em></u><em> :</em>

Let 4x + 2y = 14 be the first equation.

Let 2x + y = 7 be the second equation.

The solution (x,y) can be determined by solving the two equations, if only the given two equations are different.

<u><em>step 3</em></u><em> :</em>

In first equation taking 2 out as a common factor on both sides, the equation becomes:

2 (2x + y) = 2 (7)

So, the first equation resembles the second one.

<u><em>step 4</em></u><em> :</em>

Since both the equations are similar, they cannot be solved to get a solution. Therefore, the system of equation has no solution which is also known as inconsistent.

5 0
3 years ago
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