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2 years ago

6

Determine the sum of the following series. ∑n=1[infinity](4^n+8^n/13n)

1 answer:

Gnoma [55]2 years ago

3 0

N=1 4+7n=71 that is the answer

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Definition of exponential?

lutik1710 [3]

Answer:

(of an increase) becoming more and more rapid.

"the social security budget was rising at an exponential rate"

Step-by-step explanation:

✨research ✨

4 0

2 years ago

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A plane flies with an average velocity of -98.5 m/s for 45.0s .What is the displacement?

Gnesinka [82]

Answer:

-4432.5 m

Step-by-step explanation:

Displacement is measured in meters, so will be the product of velocity in m/s and time in s.

(-98.5 m/s)×(45.0 s) = -4432.5 m

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If you're concerned with significant figures, you can round this to -4430 m, which has the required 3 significant figures.

8 0

2 years ago

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A cylinder has a height of 14 inches and a radius of 5 inches. What is its volume? Use ≈ 3.14 and round your answer to the nea

lianna [129]

Answer:

1099 inches

Step-by-step explanation:

We know that formula for volume of a cylinder is:

$V=\pi r^2h$

$V=\pi 5^2*14$

$V=\pi 25*14$

$V=\pi 350$

$V=3.14*350$

$V=1099$

8 0

2 years ago

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Write anwsers as fractions using / as fraction bar and write the greater value first

puteri [66]

Answer:

Final answer is $\frac{3}{2}$ or $\frac{1}{2}$ .

Step-by-step explanation:

Given equation is $\log_x\left(8x-3\right)-\log_x\left(4\right)=2$

Now we need to solve that equation for x.

$\log_x\left(8x-3\right)-\log_x\left(4\right)=2$

Apply formula $\log_c\left(A\right)-\log_c\left(B\right)=\log_c\left(\frac{A}{B}\right)$

$\log_x\left(\frac{8x-3}{4}\right)=2$

Apply formula $\log_c\left(A\right)=b \Rightarrow c^b=A$

$x^2=\frac{8x-3}{4}$

$4x^2=8x-3$

$4x^2-8x+3=0$

$4x^2-2x-6x+3=0$

$2x(2x-1)-3(2x-1)=0$

$(2x-3)(2x-1)=0$

$2x-3=0$ or $2x-1=0$

$2x=3$ or $2x=1$

$x=\frac{3}{2}$ or $x=\frac{1}{2}$

Hence final answer is $\frac{3}{2}$ or $\frac{1}{2}$ .

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3 years ago

Can someone please help me? You get points and it makes me happy.

Sergio039 [100]

The answer is C. $9.00

7 0

2 years ago

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