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2 years ago

Work

Mathematics

1 answer:

liubo4ka [24]2 years ago

7 0

Answer:

Step-by-step explanation:

the quotient of is the result of dividing two terms

The first term is - Twice a number t: 2t

The second term is - and 12: 12 2t÷12or 2t12

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2. You and six friends order three large pizzas for a pizza party. Each large pizza had twelve pieces. There

Veronika [31]

Answer:

Step-by-step explanation:

12+12+12= 36

36-7= 29

3+3+3=9

29-9=20

20÷4=5

7 0

2 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

3 years ago

I need the steps and answers pleass

Schach [20]

Answer:

hope this helps you

have a great day

5 0

2 years ago

A moving company charges $30 plus $0.15 Per mile to rent a moving van another company charges $15 plus $0.20 per mile to rent th

Ratling [72]

30 + x0.15=y is the first equation
15+x0.20=y is the second equation
Where in both x equals the amount of days it takes and y is your out come
You can solve this problem two ways one by finding the common denominator or just multiply x until you get the same awnser between the both of them

6 0

3 years ago

write an equation in general form of the line passing through (3, -5) whose slope is the negative reciprocal of -1/4

IceJOKER [234]

Negative reciprocal means
-1/4 times x=-1 solve for x
multiply both sides by -4/1
x=-4
so slope=-4
y=mx+b
m=slope, b=y intercpet
y=-4x+b
input 3 for x and -5 for y
-5=-4(3)+b
-5=-12+b
add 12 to both sides
7=b
the equation in slope intercept form is y=-4x+7
that is my standard form
another standard form is
y+4x=7

4 0

2 years ago