All categories

3 years ago

Write the following inequality in slope-intercept form. −8x + 4y ≥ −52

2 answers:

6 0

Slope-intercept form:
y=mx+b

-8x+4y≥-52
Add 8x for both side
-8x+8x+4y≥-52+8x
4y≥8x-52
Divided 4 for both side
4y/4≥(8x-52)/4
y≥2x-13. Hope it help!

Shkiper50 [21]3 years ago

4 0

-8x+4y≥-52

Add 8x to both sides:
-8x+8x+4y≥8x-52

Now we have:
4y≥8x-52

Divide everything by 4.

Final answer:
y≥4x-13

You might be interested in

Can someone Please help with this

Mekhanik [1.2K]

answers for the blank: 23 and 25

explanation: plug in the n to the equation and solve! :)

hope this helps ❤ from peachimin

5 0

3 years ago

Write an equation of the line that passes through the points (2,−1) and (0,1).

Alina [70]

0,1 is the y-intercept

5 0

3 years ago

Read 2 more answers

PLEASE HELP I DO NOT UNDERSTAND AT ALL ITS PRECALC PLEASE SERIOUS ANSWERS

Elina [12.6K]

You want to end up with $A\sin(\omega t+\phi)$ . Expand this using the angle sum identity for sine:

$A\sin(\omega t+\phi)=A\sin(\omega t)\cos\phi+A\cos(\omega t)\sin\phi$

We want this to line up with $2\sin(4\pi t)+5\cos(4\pi t)$ . Right away, we know $\omega=4\pi$ .

We also need to have

$\begin{cases}A\cos\phi=2\\A\sin\phi=5\end{cases}$

Recall that $\sin^2x+\cos^2x=1$ for all $x$ ; this means

$(A\cos\phi)^2+(A\sin\phi)^2=2^2+5^2\implies A^2=29\implies A=\sqrt{29}$

Then

$\begin{cases}\cos\phi=\frac2{\sqrt{29}}\\\sin\phi=\frac5{\sqrt{29}}\end{cases}\implies\tan\phi=\dfrac{\sin\phi}{\cos\phi}=\dfrac52\implies\phi=\tan^{-1}\left(\dfrac52\right)$

So we end up with

$2\sin(4\pi t)+5\cos(4\pi t)=\sqrt{29}\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)$

8 0

3 years ago

Read 2 more answers

Convert 140° from degree measure to radians

larisa86 [58]

Answer:

7pi/9

Step-by-step explanation:

140 degrees * (pi/180 degrees) = 7pi/9

3 0

3 years ago

What is the quotient of the division below?

erastovalidia [21]

Answer:

C. 48

Step-by-step explanation:

336/7 is 48

3 0

3 years ago

Read 2 more answers