Answer:
Solution given:
we have
m<PTQ=m<RTS
x+20=3x+12.
20-12=3x-x
x=

now
16.m<PTQ=x+20=4+20=24°
17.
again
m<PTR+m<PTQ=180°[supplementary]
so
m<PTR=180°-24°<u>=1</u><u>5</u><u>6</u><u>°</u><u> </u><u>i</u><u>s</u><u> </u><u>your</u><u> </u><u>answer</u>
Answer:

The domain for x is all real numbers greater than zero and less than 5 com
Step-by-step explanation:
<em><u>The question is</u></em>
What is the volume of the open top box as a function of the side length x in cm of the square cutouts?
see the attached figure to better understand the problem
Let
x -----> the side length in cm of the square cutouts
we know that
The volume of the open top box is

we have



substitute

Find the domain for x
we know that

so
The domain is the interval (0,5)
The domain is all real numbers greater than zero and less than 5 cm
therefore
The volume of the open top box as a function of the side length x in cm of the square cutouts is

Answer:
radius = 10
Step-by-step explanation:
As AB is the diameter of the circle, the angle ACB inscribes an arc of 180 degrees, so it is a 90 degrees angle, then the triangle ACB is a right triangle, with hypotenusa AB.
We can find the length of the diameter AB using the Pythagoras' theorem:
AB^2 = CA^2 + CB^2
AB^2 = 12^2 + 16^2 = 400
AB = 20
The radius of the circle is half the diameter, so the radius is 20 / 2 = 10
12(pi)
diameter is 4, so radius should be 2
closest height to equal 12(pi) would be 0.96
Okay so this is how I got the answer.... I didn’t lol