The formula for the nth term of a geometric sequence:
a₁ - the first term, r - the common ratio
![54, a_2, a_3, 128 \\ \\ a_1=54 \\ a_4=128 \\ \\ a_n=a_1 \times r^{n-1} \\ a_4=a_1 \times r^3 \\ 128=54 \times r^3 \\ \frac{128}{54}=r^3 \\ \frac{128 \div 2}{54 \div 2}=r^3 \\ \frac{64}{27}=r^3 \\ \sqrt[3]{\frac{64}{27}}=\sqrt[3]{r^3} \\ \frac{\sqrt[3]{64}}{\sqrt[3]{27}}=r \\ r=\frac{4}{3}](https://tex.z-dn.net/?f=54%2C%20a_2%2C%20a_3%2C%20128%20%5C%5C%20%5C%5C%0Aa_1%3D54%20%5C%5C%0Aa_4%3D128%20%5C%5C%20%5C%5C%0Aa_n%3Da_1%20%5Ctimes%20r%5E%7Bn-1%7D%20%5C%5C%0Aa_4%3Da_1%20%5Ctimes%20r%5E3%20%5C%5C%0A128%3D54%20%5Ctimes%20r%5E3%20%5C%5C%0A%5Cfrac%7B128%7D%7B54%7D%3Dr%5E3%20%5C%5C%20%5Cfrac%7B128%20%5Cdiv%202%7D%7B54%20%5Cdiv%202%7D%3Dr%5E3%20%5C%5C%0A%5Cfrac%7B64%7D%7B27%7D%3Dr%5E3%20%5C%5C%0A%5Csqrt%5B3%5D%7B%5Cfrac%7B64%7D%7B27%7D%7D%3D%5Csqrt%5B3%5D%7Br%5E3%7D%20%5C%5C%0A%5Cfrac%7B%5Csqrt%5B3%5D%7B64%7D%7D%7B%5Csqrt%5B3%5D%7B27%7D%7D%3Dr%20%5C%5C%0Ar%3D%5Cfrac%7B4%7D%7B3%7D)
Answer:
0.6 is your answer if you need explanation than comment me
Setting AC = BD because the diagonals are congruent, then 3y/5 = 3y -4
y=5/3
So then AC = 1 = BD
Answer:
2472 students
Step-by-step explanation:
<u>What we know:</u>
- There are 6 schools
- Each school has 412 students
<u>What we're trying to solve:</u>
How many students are there altogether?
To find this, we multiply 412 by 6. Doing this is essentially the same as adding 412 to itself 6 times for each school there is in the district.
412 students × 6 schools
= 2472 students
Therefore, there are 2472 students in the district.
I hope this helps!
