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3 years ago

WILL GIVE BRAINIEST PLEASE ANSWER AND GIVE EXPLANATION THANKS

Mathematics

1 answer:

solniwko [45]3 years ago

4 0

Answer:

I just have a question how do you give brainiest?

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Anastasy [175]

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3 years ago

Help with this plz? <br>Topic (Similarity)

Darina [25.2K]

Answer:

MATH

Step-by-step explanation:

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3 years ago

Cam was bom weighing 8.6 pounds. Every day, he gained the same amount of weight. After 30 days, he weighed 14.9

guajiro [1.7K]

Answer:

So I'm not entirely sure about this one, but I believe he gained 0.21 pounds each day for 30 days

Step-by-step explanation:

Since Cam was born weighing 8.6 pounds and after 30 days weighed 14.9, subtract 14.9-8.6 to find how much he gained.

14.9 - 8.6 = 6.3

Now that we know how much he gained, we have to find out how much he gained per day for 30 days. To do this, divide 6.3 by 30

6.3 / 30 = 0.21

To make sure this answer is correct, multiply 0.21 by 30 and add that to 8.6. It should equal 14.9.

I hope this helps!

7 0

3 years ago

Read 2 more answers

I will give you the brainliest if you answer asap please

IRINA_888 [86]

Answer:

168

Step-by-step explanation:

You just need to triple the answer to get 168.

8 0

3 years ago

This is a question on my partial fractions homework, but no matter what I try I can't figure it out..

Ierofanga [76]

$\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{a_1x+a_0}{(x+1)^2}+\dfrac b{x+2}$
$\implies\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{(a_1x+a_0)(x+2)+b(x+1)^2}{(x+1)^2(x+2)}$
$\implies x^2+x+1=(a_1+b)x^2+(2a_1+a_0+2b)x+(2a_0+b)$
$\implies\begin{cases}a_1+b=1\\2a_1+a_0+2b=1\\2a_0+b=1\end{cases}\implies a_1=-2,a_0=-1,b=3$

So you have

$\displaystyle\int_0^2\frac{x^2+x+1}{(x+1)^2(x+2)}\,\mathrm dx=-2\int_0^2\frac x{(x+1)^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$
$=\displaystyle-2\int_1^3\dfrac{x-1}{x^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$

where in the first integral we substitute $x\mapsto x+1$ .

$=\displaystyle-2\int_1^3\left(\frac1x-\frac1{x^2}\right)\,\mathrm dx-\frac1{1+x}\bigg|_{x=0}^{x=2}+3\ln|x+2|\bigg|_{x=0}^{x=2}$
$=-2\left(\ln|x|+\dfrac1x\right)\bigg|_{x=1}^{x=3}-\dfrac23+3(\ln4-\ln2)$
$=-2\left(\ln3+\dfrac13-1\right)-\dfrac23+3\ln2$
$=\dfrac23+\ln\dfrac89$

4 0

3 years ago