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2 years ago

Write an equation that you can use to solve for x. Enter your answer in the box.

Mathematics

2 answers:

Sindrei [870]2 years ago

7 0

Answer:

x+63 = 90

Step-by-step explanation:

x and 63 add to a right angle so their sum is 90

x+63 = 90

We can solve

x = 90-63

x = 27

N76 [4]2 years ago

5 0

The answer is X=27 degrees

Steps-
90-63=27

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The percentage of people under the age of 18 was 23.5% in New York City, 25.8% in Chicago, and 26% in Los Angeles.

Hatshy [7]

Answer:

0.0158 = 1.58% probability that all of them are under 18.

Step-by-step explanation:

Probability of independent events:

If three events, A, B and C are independent, the probability of all happening is the multiplication of the probabilities of each happening, that is:

$P(A \cap B \cap C) = P(A)P(B)P(C)$

The percentage of people under the age of 18 was 23.5% in New York City, 25.8% in Chicago, and 26% in Los Angeles.

This means that $P(A) = 0.235, P(B) = 0.258, P(C) = 0.26$

If one person is selected from each city, what is the probability that all of them are under 18?

Since the three people are independent:

$P(A \cap B \cap C) = 0.235*0.258*0.26 = 0.0158$

0.0158 = 1.58% probability that all of them are under 18.

8 0

2 years ago

aaron walks 2 1/8 miles to his friends house. then, they walk 5/6 miles to the park .finally, aaron walks 1 3/4 miles to get bac

lisov135 [29]

Answer:did you check and see if anybody asked this question before??

Step-by-step explanation:

4 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 $x\mapsto x+1$ .

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