A point on a wheel moves at 32π/9 radians per minute. How many seconds are required for the point to complete 7 revolutions?

Which right prism would have the same volume as a square prism with a base area of 36 m2 and a height of 3 m?

Neko [114]

Well to find the volume of the square prism, first you need to multiply the base area (36) by its height (3) and you would end up with 108m3 as the volume. You may not be familiar with this because u are most likely taught length times width times height as volume, but what i just did is the same because length times width is the sum of the base area, and height was already given. Anyways, now that we have gotten the square prism volume (108m3) we need to find which prism has the same volume. I’m gonna save u from searching and tell you that it is the rectangular prism because length times width times height (18x2x3) equaled 108 just like the square prism.

8 0

3 years ago

Read 2 more answers

Robert drew a parallelogram with 2 55 degree angles. What are the measures of the 2 other angles? Both of the angles have to be

Licemer1 [7]

Angles in a parallelogram have to equal 360.

Let x represent the measurement of the unknown angle.

2(55) + 2x = 360
110 + 2x = 360
2x = 250
x = 125

125 degrees

5 0

3 years ago

Which polynomial, when added to the polynomial 5x^2–3x–9, is equivalent to:<br><br> x^2–5x+6?

inn [45]

Look at the attachment

8 0

3 years ago

How would I figure out a linear equation with two plot points with fractions? (1/3(6),3) and (1/2(19),-2) are my two plot points

murzikaleks [220]

Answer: $m=\frac{-30}{79},\ y=5.40$

Step-by-step explanation:

Given

Points are $\left(6\frac{1}{3},3\right),\ \left(19\frac{1}{2},-2\right)$

Convert mixed numeral to fraction

$\left(6\frac{1}{3},3\right)\Rightarrow \left(\dfrac{19}{3},3\right)\\\\\left(19\frac{1}{2},-2\right)\Rightarrow \left(\dfrac{39}{2},-2\right)$

Slope of the line is

$\Rightarrow m=\dfrac{3-(-2)}{\dfrac{19}{3}-\dfrac{39}{2}}\\\\\Rightarrow m=\dfrac{5}{\dfrac{38-117}{6}}\\\\\Rightarrow m=\dfrac{-30}{79}$

Equation of line

$\Rightarrow \dfrac{y-3}{x-\dfrac{19}{3}}=\dfrac{-30}{79}\\\\\text{for y-intercept, put x=0}\\\\\Rightarrow \dfrac{y-3}{0-\dfrac{19}{3}}=\dfrac{-30}{79}\\\\\Rightarrow y-3=\dfrac{190}{79}\\\\\Rightarrow y=3+2.40\\\Rightarrow y=5.40$