Ask question

Ask question

All categories

3 years ago

8

What does each variable in the Slope-Intercept Form equation represent?

1 answer:

Virty [35]3 years ago

7 0

Y represent the y axis m represents the rate of change x represents the x axis b represents the total

You might be interested in

A cats' home has a raffle to raise money.

Dominik [7]

An answer obviously

3 0

3 years ago

Theses are the only questions I have left PLZZ HELPPPPP

Ksenya-84 [330]

Answer:

Step-by-step explanation:

1.

Dependent variable: a

Independent variable: c

2.

(in this order)

x = 9

x = 15

x = 4

x = 36 (assuming that says x = 12 * 3)

6 0

3 years ago

Read 2 more answers

Can you prove it??<br>it's hard,I tried but couldn't solve this.

BARSIC [14]

I'll abbreviate $s=\sin\theta$ and $c=\cos\theta$ , so the identity to prove is

$\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1s\right)$

On the left side, we can simplify a bit:

$\dfrac{s+c+1}{s+c-1}=\dfrac{s+c-1+2}{s+c-1}=1+\dfrac2{s+c-1}$

$\dfrac{1+s-c}{1-s+c}=-\dfrac{-2+1-s+c}{1-s+c}=-1+\dfrac2{1-s+c}$

Then

$\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1{s+c-1}-\dfrac1{1-s+c}\right)$

So the establish the original equality, we need to show that

$\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac1s$

Combine the fractions:

$\dfrac{(1-s+c)-(s+c-1)}{(s+c-1)(1-s+c)}=\dfrac{2-2s}{c^2-s^2+2s-1}$

We can rewrite the denominator as

$c^2-s^2+2s-1=c^2+s^2-2s^2+2s-1$

then using the fact that $c^2+s^2=\cos^2\theta+\sin^2\theta=1$ , we get

$1-2s^2+2s-1=2s-2s^2$

so that we have

$\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac{2-2s}{2s-2s^2}=\dfrac1s$

as desired.

6 0

4 years ago

The vertex of this parabola is at (4,-3). which of the following could be its equation?

Mama L [17]

The answer is A.
I used the formula for finding the vertex of a parabola and I got the first answer choice. So A is correct. :)

6 0

4 years ago

Read 2 more answers

For the function given below, find a formula for the riemann sum obtained by dividing the interval [a,b] into n equal subinter

Nata [24]

We split [2, 4] into $n$ subintervals of length $\dfrac{4-2}n=\dfrac2n$ ,

$[2,4]=\left[2,2+\dfrac2n\right]\cup\left[2+\dfrac2n,2+\dfrac4n\right]\cup\left[2+\dfrac4n,2+\dfrac6n\right]\cup\cdots\cup\left[2+\dfrac{2(n-1)}n,4\right]$

so that the right endpoints are given by the sequence

$x_i=2+\dfrac{2i}n=\dfrac{2(n+i)}n$

for $1\le i\le n$ . Then the Riemann sum approximating

$\displaystyle\int_2^42x\,\mathrm dx$

is

$\displaystyle\sum_{i=1}^nf(x_i)\dfrac{4-2}n=\frac8{n^2}\sum_{i=1}^n(n+i)=\frac8{n^2}\left(n^2+\frac{n(n+1)}2\right)=\frac{12n+4}n$

The integral is given exactly as $n\to\infty$ , for which we get

$\displaystyle\int_2^42x\,\mathrm dx=\lim_{n\to\infty}\frac{12n+4}n=12$

To check: we have

$\displaystyle\int_2^42x\,\mathrm dx=x^2\bigg|_2^4=4^2-2^2=16-4=12$

7 0

3 years ago