I believe the answer would be 1.17
Answer:the potato is ugly
Step-by-step explanation:because it has marks
Okay, first let's look at what they give us!
The measure of angle 2 is 3x + 1
Measure of angle 3 is 2x + 4
And they give us that one angle is right which means it is 90°
Now if we use what we know of triangles, we know that all the angles in a triangle add up to equal 180 which means if we add angle 2, angle 3, and the right angle together we should get 180. Let's write an equation for it:
3x + 1 + 2x + 4 + 90 = 180
First we will add together liked terms!
3x + 2x = 5x
1 + 4 + 90 = 95
This gets us:
5x + 95 = 180
Second, let's get rid of that 95 by subtracting it from both sides, after doing this it should leave us with:
5x = 85
Third we need to get the x by itself and we can do it by dividing both sides by 5 to get:
x = 17
Now the question asks to find the measure of angle 2, given that angle 2 is 3x + 1 all there is left to do is to plug in 17 for x!
3(17) + 1
51 + 2 to get us 52!
Answer: 52
<h2>
Answer:</h2>
<u>First Part</u>
Given that

We have that

<u>Second Part</u>
Given that

If the Diameter were reduced by half we have that

This shows that the volume would be
of its original volume
<h2>
Step-by-step explanation:</h2>
<u>First Part</u>
Gather Information


Calculate Radius from Diameter

Use the Radius on the Volume formula

Before starting any calculation, we try to simplify everything we can by expanding the exponent and then factoring one of the 9s

We can see now that one of the 3s can be already divided by the 3 in the denominator

Finally, since we can't simplify anymore we just calculate it's volume


<u>Second Part</u>
Understanding how the Diameter reduced by half would change the Radius

Understanding how the Radius now changes the Volume

With the original Diameter, we have that

If the Diameter were reduced by half, we have that

But we can see that the numerator is exactly the original Volume!
This shows us that the Volume would be
of the original Volume if the Diameter were reduced by half.