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defon
3 years ago
15

Have to fine the answer to u 15>-4 (3-u)+5u

Mathematics
1 answer:
Galina-37 [17]3 years ago
3 0
I hope this helps you



15>-12+4u+5u


27>9u


3>u
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Estimate the quotient of 239.63 ÷ 7.51.
Crank

Answer:

About 30

Step-by-step explanation:

The value 239.63 rounds to 240

The value 7.51 rounds to 8

Divide 240 over 8 and we get 240/8 = 30

6 0
2 years ago
Another math problem that I need help on
leonid [27]
The answer is 0.4 because there is a 4/10 chance of an odd number between 3 and 10 being chosen out of the numbers listed.
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Use the Integral Test to determine whether the series is convergent or divergent
Inga [223]

Answer:

A. \sum_{n=1}^{\infty}\frac{n}{e^{15n}} converges by integral test

Step-by-step explanation:

A. At first we need to verify that the function which the series is related (\frac{n}{e^{15n}}) fills the necessary conditions to ensure that the test is effective.

*f(x) must be continuous or differentiable

*f(x) must be positive and decreasing

Let´s verify that f(x)=\frac{n}{e^{15n}} fills these conditions:

*Considering that eˣ≠0 for all x, the function f(x)=\frac{n}{e^{15n}} does not have any discontinuities, so it´s continuous

*Because eˣ is increasing:

      if a<b ,then eᵃ<eᵇ

      if 0<eᵃ<eᵇ ,then 1/eᵃ > 1/eᵇ

      if 1/eᵃ > 1/eᵇ and a<b, then a/eᵃ<b/eᵇ

  We conclude that f(x)=\frac{n}{e^{15n}} is decreasing

*Because eˣ is always positive and the sum is going from 1 to ∞, this show that f(x)=\frac{n}{e^{15n}} is positive in [1,∞).

Now we are able to use the integral test in f(x)=\frac{n}{e^{15n}} as follows:

\sum_{n=1}^{\infty}\frac{n}{e^{15n}}\ converges\ \leftrightarrow\ \int_{1}^{\infty}\frac{x}{e^{15x}}\ dx\ converges

Let´s proceed to integrate f(x) using integration by parts

\int_{1}^{\infty}\frac{x}{e^{15x}}\ dx=\int_{1}^{\infty}xe^{-15x}\ dx

Choose your U and dV like this:

U=x\ \rightarrow dU=1\\ dV=e^{-15x}\ \rightarrow V=\frac{-e^{-15x}}{15}

And continue using the formula for integration by parts:

\int_{1}^{\infty}Udv = UV|_{1}^{\infty} - \int_{1}^{\infty}Vdu

\int_{1}^{\infty}xe^{-15x}\ dx= \frac{-x}{15e^{15x}}|_{1}^{\infty} -\frac{-1}{15} \int_{1}^{\infty}e^{-15x}\ dx

\int_{1}^{\infty}xe^{-15x}\ dx= \frac{-x}{15e^{15x}}|_{1}^{\infty} -\frac{-1}{15}(\frac{-1}{15e^{15x}})|_{1}^{\infty}

\int_{1}^{\infty}xe^{-15x}\ dx= \frac{-x}{15e^{15x}}|_{1}^{\infty} -\frac{1}{225e^{15x}}|_{1}^{\infty}

Because we are dealing with ∞, we´d rewrite it as a limit that will help us at the end of the integral:

\int_{1}^{\infty}xe^{-15x}\ dx= \lim_{b \to{\infty}}(\frac{-x}{15e^{15x}}|_{1}^{b}-\frac{1}{225e^{15x}}|_{1}^{b})

\int_{1}^{\infty}xe^{-15x}\ dx= \lim_{b \to{\infty}} \frac{-b}{15e^{15b}}-\frac{1}{225e^{15b}}-(\frac{-1}{15e^{15}}-\frac{1}{225e^{15}})

\int_{1}^{\infty}xe^{-15x}\ dx= ( \lim_{b \to{\infty}} \frac{-b}{15e^{15b}}-\frac{1}{225e^{15b}})+\frac{1}{15e^{15}}(1-\frac{1}{15})

We only have left to solve the limits, but because b goes to  ∞ and it is in an exponential function on the denominator everything goes to 0

\lim_{b \to{\infty}} \frac{-b}{15e^{15b}}-\frac{1}{225e^{15b}} = 0

\int_{1}^{\infty}xe^{-15x}\ dx= \frac{1}{15e^{15}}(1-\frac{1}{15})

Showing that the integral converges, it´s the same as showing that the series converges.

By the integral test \sum_{n=1}^{\infty}\frac{n}{e^{15n}} converges

7 0
2 years ago
Please hurry ASAP One ride on a city bus costs $1.50. Martina has $18 on her bus pass. Write and solve an equation to find how m
bogdanovich [222]

Answer:

She can ride the bus 12 times before she needs to load more money on  her pass.

Step-by-step explanation

1. You would take 18 and divide it by 1.5 (1.50) to get 12.

6 0
2 years ago
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Can someone answer this please?
Wittaler [7]
V=pi (r^2) h/3
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3 0
3 years ago
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