Ok, so for this you're likely graphing in y=mx+b format (slope intercept). So to start, you'all first subtract 3x from the equation to get 6y=-3x+16. Now you'll divide the equation by 6 to get y alone. Your final equation will be y=-1/2x+16/6

7 0

2 years ago

Mike just hopped on the edge of a merry-go-round. What are his linear and angular speeds if the diameter of the merry-go-round i

dsp73

Step-by-step explanation:

Given that,

The diameter of the merry-go-round, d = 14 feet

Time taken, t = 6 seconds

Radius, r = 7 feet

The linear speed of the merry-go-round is given by :

$v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 7}{6}\\\\=7.33\ m/s$

Also,

$v=r\omega$

Where

$\omega$ is the angular speed

So,

$\omega=\dfrac{v}{r}\\\\\omega=\dfrac{7.33}{7}\\\\=1.04\ rad/s$

Hence, his linear and angular speeds are 7.33 m/s and 1.04 rad/s.

8 0

3 years ago

Susan had a $10 coupon for the purchase of any time item. She bought a coat that was on sale for

gulaghasi [49]

The original price is 260

2*125+10 please mark me brainliest

7 0

3 years ago