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3 years ago

Hellppp meeeeeee PLEASEEEEE.....

Mathematics

2 answers:

sleet_krkn [62]3 years ago

4 0

Answer:

2+2

Step-by-step explanation:

what you will do is add 2 to the other to and get 4

densk [106]3 years ago

4 0

Hm try photo math THATS what helps me

You might be interested in

Help soon! Its very late at night for me ;-;

stich3 [128]

—
Max: 12
Median: 7
LQ: 1, 2
UQ: 11, 12

5 0

3 years ago

Read 2 more answers

ASAP

vagabundo [1.1K]

Answer:

$\frac{x+12}{4(x+3)}$

Step-by-step explanation:

we have

$(\frac{3}{x} +\frac{1}{4}):(1+\frac{3}{x})$

Multiply by 4x numerator and denominator to remove the fractions

$(\frac{3}{x}(4x) +\frac{1}{4}(4x)):(1(4x)+\frac{3}{x}(4x))$

$\frac{12+x}{4x+12}$

Factor 4 in the denominator

$\frac{x+12}{4(x+3)}$

7 0

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 $b=-1$ .

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

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

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

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

4 years ago

Write an equation in the line in slope intercept form

Sergio039 [100]

Answer:

Y=2x-1

Step-by-step explanation:

In order to find the slope, its rise over run. So you would move from (0,-1) to the next point it intercepts, (1,1) and find it from there

As for the Y-Intercept, it would just be wherever the line connects with the Y axis, which happens to be (0,-1) or Negative 1.

8 0

3 years ago

Find the distance between the points A and B given below.

GalinKa [24]

Answer:

use formula

Step-by-step explanation:

distance between the points A and B=

√((x2-x1)²+(y2-y1)²)

5 0

3 years ago