Answer:
When the diameter is given you just take half of it so simply put what is half of 16pi and it will give your answer. :)
Step-by-step explanation:
Answer:
She subtracted 7x instead of dividing it. Personally, i wouldn't have done that. I would subtract the 14 first, then divide.
Answer:
<h2><em><u>
A to C = 25
</u></em></h2><h2><em><u>
A to B = 13
</u></em></h2><h2><em><u>
C to B = 37
</u></em></h2><h2><em><u>
</u></em></h2>
Step-by-Step Explanation:
<em><u>Perimeter</u></em> = 75
<em><u>Sides:</u></em>
2x + 3
3x + 4
2x - 9
<h2 /><h2><em><u>
1. Equal the sides added together to the perimeter</u></em></h2>
75 = 2x + 3 + 3x + 4 + 2x - 9
<h2><em><u>
2. Simplify Like terms</u></em></h2>
2x + 3 + 3x + 4 + 2x - 9 = 7x - 2
<h2><em><u>
3. Place the equation back together</u></em></h2>
75 = 7x - 2
<h2><em><u>
4. Isolate the variables and numbers</u></em></h2>
75 = 7x - 2
+2 +2
77 = 7x
<h2><em><u>
5. Simplify the equation</u></em></h2>
77 = 7x
/7 /7
<h2><em><u>
11 = x
</u></em></h2>
<h2><em><u>
6. Substitute the value of x into the side lengths.</u></em></h2>
2x + 3 = 2(11) + 3 = 22 + 3 = <em><u>25</u></em>
3x + 4 = 3(11) + 4 = 33 + 4 = <em><u>37</u></em>
2x - 9 = 2(11) - 9 = 22 - 9 = <em><u>13</u></em>
Answer:
![a_5 = 32](https://tex.z-dn.net/?f=a_5%20%3D%2032)
Step-by-step explanation:
The nth term for the geometric sequence is given by:
![a_n = a_1 \cdot r^{n-1}](https://tex.z-dn.net/?f=a_n%20%3D%20a_1%20%5Ccdot%20r%5E%7Bn-1%7D)
where,
is the first term
r is the common ratio
n is the number of terms.
As per the statement:
For the geometric sequence of
and r=2
We have to find ![a_5](https://tex.z-dn.net/?f=a_5)
for n = 5;
![a_5=a_1 \cdot r^{n-1}](https://tex.z-dn.net/?f=a_5%3Da_1%20%5Ccdot%20r%5E%7Bn-1%7D)
Substitute the given values we have;
![a_5 = 2 \cdot 2^4 = 2 \cdot 16](https://tex.z-dn.net/?f=a_5%20%3D%202%20%5Ccdot%202%5E4%20%3D%202%20%5Ccdot%2016)
⇒![a_5 = 32](https://tex.z-dn.net/?f=a_5%20%3D%2032)
Therefore, the value of
is, 32
Quotients can sometimes be irrational so it is necessary to estimate them, products on the other hand are only rational, provided they aren't multiplied by an irrational, so it is not necessary to estimate