What is the multiplier for the table below? (Write multiplier as x.x)

Mathematics

1 answer:

chubhunter [2.5K]3 years ago

8 0

The average rate of change is m = 5.934

Hope this is what you were looking for.

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PLEASE HELP ME Factor each expression, 2n^(2)+ 3n -2

kotegsom [21]

I would suggest factoring using the grouping method. The answer I got was (2n-1)(n+2)

5 0

4 years ago

Show work and explain with formulas.

Juli2301 [7.4K]

20 Answer: a₁ = 4

Step-by-step explanation:

$a_n=324,\ r=3,\ n=5\\\\a_n=a_1 \cdot r^{n-1}\\\\324=a_1\cdot 3^{5-1}\\\\\dfrac{324}{3^4}=a_1\\\\\dfrac{324}{81}=a_1\\\\\large\boxed{4}=a_1$

21 Answer: n = 13

Step-by-step explanation:

$a_n=\dfrac{1}{64},\ a_1=64,\ r=\dfrac{1}{2}\\\\a_n=a_1 \cdot r^{n-1}\\\\\dfrac{1}{64}=64\cdot \bigg(\dfrac{1}{2}\bigg)^{n-1}\\\\\dfrac{1}{64\cdot 64}=\bigg(\dfrac{1}{2}\bigg)^{n-1}\\\\\dfrac{1}{2^6\cdot 2^6}=\dfrac{1}{2^{n-1}}\\\\6+6=n-1\\\\\large\boxed{13}=n$

22 Answer: n = 5

Step-by-step explanation:

$a_n=48,\ a_1=1875\ r=\dfrac{2}{5}\\\\a_n=a_1 \cdot r^{n-1}\\\\48=1875\cdot \bigg(\dfrac{2}{5}\bigg)^{n-1}\\\\\dfrac{48}{1875}=\bigg(\dfrac{2}{5}\bigg)^{n-1}\\\\\dfrac{16}{625}=\bigg(\dfrac{2}{5}\bigg)^{n-1}}\\\\\bigg(\dfrac{2}{5}\bigg)^4=\bigg(\dfrac{2}{5}\bigg)^{n-1}}\\\\4=n-1\\\\\large\boxed{5}=n$