What is the sum of the geometric series in which a1 = 7, r = 3, and an = 1,701?

1 answer:

4 0

Answer: The sum of the series is 2,548.

In a geometric series, the values are being multiplied by a constant rate. In this series, we are started at 7 and multiplying by 3.

Our series would be:
7 21 63 189 567 1701

If we add up those values, we have a total of 2,548.

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A case of water costs $3.84 and contains 24 bottles of water. Determine the cost of each bottle of water.

Darya [45]

Answer:

0.16

Step-by-step explanation:

3.84/24=0.16

Divide the total price by the amount of waterbottles in the case to find each individual one.

Hope this helps!

3 0

3 years ago

Find CD if C(0,3) and D(4,7)

weeeeeb [17]

$C(0,3) \\
x_1=0 \\
y_1=3 \\ \\
D(4,7) \\
x_2=4 \\
y_2=7 \\ \\
\overset{\underline{\ \ \ \ }}{CD} =\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(4-0)^2+(7-3)^2}= \\
=\sqrt{4^2+4^2}=\sqrt{16+16}=\sqrt{16 \times 2}=4\sqrt{2} \\ \\
\boxed{\overset{\underline{\ \ \ \ }}{CD} = 4\sqrt{2}}$

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3 years ago

Will mark Brainlest hellpppp

Anna007 [38]

$h(-3) - h( - 2) = - \frac{5}{ 8} \\ \\$

Step-by-step explanation:

$\boxed{h(x) = \frac{2 {x}^{2} - x + 1}{3x - 2}}$

$h(2) = \frac{2 {(2)}^{2} - (2) + 1}{3(2) - 2}$

$h(2) = \frac{2 {(4)} - 2 + 1}{6 - 2}$

$h(2) = \frac{ 8 - 1}{4}$

$h(2) = \frac{7}{4}$

$h( - 3) = \frac{2 {( - 3)}^{2} - ( - 3) + 1}{3( - 3) - 2} \\ h( - 3) = \frac{2 (9) + 3 + 1}{ - 9- 2} \\ h( - 3) = \frac{18 + 4}{ - 11} \\ h( - 3) = \frac{22}{ - 11} \\ h(-3) = -2$

$h( - 2) = \frac{2 {( - 2)}^{2} - ( - 2) + 1}{3( - 2) - 2} \\ h( - 2) = \frac{2 (4) + 2+ 1}{- 6 - 2} \\ h( - 2) = \frac{8 + 3}{- 8} \\ h( - 2) = \frac{11}{- 8}$

$h(3) - h( - 2) = -2 - \frac{11}{- 8} \\ h(3) - h( - 2) =-2 + \frac{11}{ 8} \\ h(3) - h( - 2) = \frac{-2}{1}+ \frac{11}{ 8} \\ h(3) - h( - 2) = \frac{-16}{8} + \frac{11}{ 8} \\ h(3) - h( - 2) = \frac{-16+11}{8} \\ h(3) - h( - 2) = \frac{-5}{8} \\ - \frac{5}{ 8}$