All categories

3 years ago

Please help me out!

Mathematics

1 answer:

Zanzabum3 years ago

8 0

Answer:

n = 7 I did 9 - 2

You might be interested in

Hiii please help i’ll give brainliest if you give a correct answer tysm!

taurus [48]

20 people

Reason - following the sequence

8 0

3 years ago

Gina drew a circle with right triangle PRQ inscribed in it, as shown below:

Darina [25.2K]

I think the correct answer is b 75

3 0

3 years ago

Read 2 more answers

write an equation of the line that's passes through the point (1,6) and is parallel to the line whose equation is y= 3x - 5

Alex Ar [27]

Parallel lines have the same slope.
so to find any line, you need a set of points, and a slope.
we already have the slope, being 3.
so, use the equation
y- (y-point)/ x-(x-point)=slope.
y-6. 3
---- = ---- (make it a fraction by puttingitover 1)
x-1. 1
cross multiply.
1(y-6)=3 (x-1)
y-6=3x-3
add 6 to both sides
Final answer: Y=3x+3

7 0

3 years ago

jason can type 275 characters in 5 minutes on his computer keyboard. at that rate, how many characters can he type in 20 minutes

Minchanka [31]

Answer:

1100 characters

Step-by-step explanation:

275 characters : 5 minutes

? characters : 20 minutes

5 x 4 = 20 minutes

275 x 4 = 1100 characters

5 0

3 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

Please help me out!​

Please help me out!