The midpoint of AB has coordinates of (4, -9). Endpoint A has coordinates (-3, -5). What are the coordinates of B?

If f(x)=-4x—2 is vertically translated 6 units up to g( x ) what is the y-intercept of g(x)

Alexxx [7]

Answer:

The y-intercept of g(x) is 4.

Step-by-step explanation:

The linear function has the equation f (x) = m*x + n where, m and n are real numbers called slope (m) and y-intercept (n).

When performing a vertical translation of a function, the graph will move from one point to another determined point in the direction of the "y" axis, that is, up or down. In this way the value of n is modified.

In the function f (x) = - 4x-2, the y-intercept has a value of -2. If this function is translated vertically 6 units up, then the new y-intercept value is -2 + 6 = 4

So, the y-intercept of g(x) is 4.

3 0

3 years ago

Read 2 more answers

How do you find the discount rate

Viktor [21]

To find the discount rate you must first, multiply the RATE by the original price. Then to find the sale price, subtract the DISCOUNT from the original price.

5 0

3 years ago

Sin(x)^4+ cos(x)^4=1/2

posledela

I'll assume that what was meant was $\sin ^4 x + \cos ^4 x = \dfrac{1}{2}$ .

The exponent in the funny place is just an abbreviation: $\sin ^4 x = (\sin x)^4$ .

I hope that's what you meant. Let me know if I'm wrong.

Let's start from the old saw

$\cos^2 x + \sin ^2x = 1$

Squaring both sides,

$(\cos^2 x + \sin ^2x)^2 = 1^2$

$\cos^4 x + 2 \cos ^2 x \sin ^2x +\sin ^4x = 1$

$\cos^4 x + \sin ^4x = 1 - 2 \cos ^2 x \sin ^2x$

So now the original question

$\sin ^4 x + \cos ^4 x = \dfrac{1}{2}$

becomes
$1 - 2 \cos ^2 x \sin ^2x = \dfrac{1}{2}$

$4 \cos ^2 x \sin ^2x = 1$

Now we use the sine double angle formula

$\sin 2x = 2 \sin x \cos x$

We square it to see

$\sin^2 2x = 4\sin^2 x \cos^2 x = 1$

Taking the square root,

$\sin 2x = \pm 1$

Not sure how you want it; we'll do it in degrees.

When we know the sine of an angle, there's usually two angles on the unit circle that have that sine. They're supplementary angles which add to $180^\circ$ . But when the sine is 1 or -1 like it is here, we're looking at $90^\circ$ and $-90^\circ$ , which are essentially their own supplements, slightly less messy.

That means we have two equations:

$\sin 2x = 1 = \sin 90^\circ$

$2x = 90^\circ + 360^\circ k \quad$ integer $k$

$x = 45^\circ + 180^\circ k$

or

$\sin 2x = -1 = \sin -90^\circ$

$2x = -90^\circ+ 360^\circ k$

$x = - 45^\circ + 180^\circ k$

We can combine those for a final answer,

$x = \pm 45^\circ + 180^\circ k \quad$ integer $k$

Check. Let's just check one, how about

$x=-45^\circ + 180^\circ = 135^\circ$

$\sin(135)= 1/\sqrt{2}$

$\sin ^4(135)=(1/\sqrt{2})^4 = 1/4$

$\cos ^4(135)=(-1/\sqrt{2})^4 = 1/4$

$\sin ^4(135^\circ) +\cos ^4(135^\circ) = 1/2 \quad\checkmark$

6 0

3 years ago

Theos mom bought an eight pack of juice boxes at the store for $4. Find the unit rate for the juice boxes.PLZ HELP ME

Otrada [13]

I believe that you need to divide 4 by 8 to get $0.50 per juice box.

3 0

4 years ago

Factor -28x +20x show work

Shalnov [3]

Let's simplify step-by-step.−28x+20x

ombine Like Terms:=−28x+20x=(−28x+20x)=−8x

Answer:=−8x

8 0

3 years ago