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)
a₁ - the first term, r - the common ratio
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Answer:
Step-by-step explanation:
slopes and equations:
find the equation thur ( 6,1 ) and (-2,-3)
find the slope m
m = (y2-y1) / (x2-x1 )
m = (-3 - 1) / (-2 -6)
m = -4 / -8
m = 1/2
now use the point-slope formula with our known slope
y-y1 = m(x-x1)
y-1 = 1/2(x-6)
y - 1 = 1/2x -3
y = 1/2x -3 +1
y = x -2
Find the equation parallel to y = 3x + 6 and thur (0,1)
Parallel means the same slope, the slope is 3 for the equation above.
use the slope-intercept formula again with the point given and the slope 3
y-1 = 3(x -0)
y - 1 = 3x
y = 3x +1
Find the equations perpendicular to 2x + y = 8 and the same y intercept as 4y = x + 3.
put both equations into proper form
y = -2x +6
y = x +
perpendicular means reciprocal slope and change the sign, the 2nd equation has an intercept of , so
y = 1/2x +
there you go Amanda :)