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3 years ago

Solve the following equation for L: P= 2L + 2W

1 answer:

3 0

P = 2L + 3w

Add -2w to both sides of the equation

2L = p - 2w

Divide both sides by 2

2L = p - 2w
/2 /2

L = 1/2P - w

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Corbin can shoot 24 baskets In 3 minutes. Which table best represents this relationship?

seraphim [82]

Answer:

Table C

Step-by-step explanation:

This is the proportional relationship:

y = kx

We have x = 3, y = 24, then k is:

24 = 3k
k = 8

The table will look as:

x = 1 ⇒ y = 1*8 = 8
x = 2 ⇒ y = 2*8 = 16
x = 5 ⇒ y = 5*8 = 40

Correct choice is C

7 0

3 years ago

Read 2 more answers

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

What is the value of -2.4 /1.5? A -1.6 B. 1.6 C. -0.9

Digiron [165]

Answer: -1.6

-2.4 divided by 1.5 is -1.6

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3 years ago

PLEASE HELP ME QUICKKK, FIRST CORRECT PERSON GETS BRAINLIEST

gladu [14]

Answer:

<h2>all a,d, and e are correct. Use the area of a circle formula to solve</h2>

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2 years ago

What is 9.01x10^3 in scientific notation

Alchen [17]

9.01 * 10^3 in scientific notation is 9.01 * 10^3.

I got this answer by moving the decimal so there is a non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent of the 10. If the decimal is binge move to the right, the exponent will be negative, but if the decimal is being moved to the left the exponent will be positive.

I hope this helps! :>

7 0

3 years ago