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Scilla [17]
3 years ago
13

Pls help many thanks

Mathematics
2 answers:
salantis [7]3 years ago
7 0

So, let's make these two values improper fractions so they're easier to work with. 4+\frac{1}{3}=\frac{12}{3}+\frac{1}{3}=\frac{13}{3} 3+\frac{1}{2}=\frac{6}{2}+\frac{1}{2}=\frac{7}{2}

So, to get the rate he grew in those months we have:

\frac{\frac{13}{3} \text{inches}}{\frac{7}{2}\text{months}} =\frac{26\text{inches}}{21\text{months}}

And to get the rate he grew in one month we have to divide both our numerator and our denominator by 21 to get:

\frac{\frac{26}{21} \text{inches}}{\text{month}}\approx \frac{1.24\text{inches}}{\text{month}} aka he grew 1.24 inches in a month.

Harlamova29_29 [7]3 years ago
6 0
1.24 inches is the best answer
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Need help question #1. Show steps please
SSSSS [86.1K]

Answer:

C

Step-by-step explanation:

We want to integrate:

\displaystyle \int\frac{4x^4+3}{4x^5+15x+2}\,dx

Notice that the expression in the denominator is quite similar to the expression in the numerator. So, we can try performing u-substitution. Let u be the function in the denominator. So:

u=4x^5+15x+2

By differentiating both sides with respect to x:

\displaystyle \frac{du}{dx}=20x^4+15

We can "multiply" both sides by dx:

du=20x^4+15\,dx

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\displaystyle \frac{1}{5}\, du=4x^4+3\,dx

Rewriting our original integral yields:

\displaystyle \int \frac{1}{4x^5+15x+2}(4x^4+3\, dx)

Substitute:

\displaystyle =\int \frac{1}{u}\Big(\frac{1}{5} \, du\Big)

Simplify:

\displaystyle =\frac{1}{5}\int \frac{1}{u}\, du

This is a common integral:

\displaystyle =\frac{1}{5}\ln|u|

Back-substitute. Of course, we need the constant of integration:

\displaystyle =\frac{1}{5}\ln|4x^5+15x+2|+C

Our answer is C.

4 0
3 years ago
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