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3 years ago

Find the value of x round to the nearest degree.

Mathematics

1 answer:

juin [17]3 years ago

6 0

Answer:

x=41.63

Step-by-step explanation:

Apply Tan

$Tan(x)=\frac{8}{9} \\x=tan^{-1}(\frac{8}{9})\\x=41.63$

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What's wrong with the problem: 5pints times 1cup divided by 2 pints = 2.5cups

lys-0071 [83]

I don't really know how the table works, but there are two cups in a pint, so 5 pints * 0.5 pints would be 2.5 pints, and 2.5 pints divided by 2 pints would be 1.25 pints?

3 0

4 years ago

Subtract fractions like 42/7- 2 1/5

dangina [55]

Answer:

3.8 in decimal form, or 3 4/5

Step-by-step explanation:

I just used a calculator, sorry :(

anyways, hope this helped! :)

6 0

3 years ago

Find the 70th term of the following arithmetic sequence.<br><br> 12,17,22,27,...

nikdorinn [45]

Since it starts at 12 and is increasing by 5 each time, the 70th term will be;
12 + 5(70)
12 + 350
362

362 is the 70th term of the sequence.

8 0

3 years ago

Read 2 more answers

Shay walks to softball practice. She leaves her home and walks 6 blocks north. She then turns east and walks 4 more blocks to th

Radda [10]

Answer:

Step-by-step explanation:

(a squared) plus (b squared) equals (c squared)

(6 squared) plus (4 squared) equals (c squared)

36+16=c squared

c equals 7

4 0

2 years ago

Read 2 more answers

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