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3 years ago

Can someone help me answer this question

Mathematics

1 answer:

OLga [1]3 years ago

6 0

Answer:

1.6

Step-by-step explanation:

2*.80=1.6

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Simplest form 4:2:8 how do you work this out

irga5000 [103]

If ur looking for the mean it would be 4.7

6 0

3 years ago

2(3x - 2) - 11x = -5x<br> ?<br> Infinitely many solutions

Alika [10]

2(3x - 2) - 11x = -5x
6x-2 - 11x=-5x
10x-2
12x

I think it right

5 0

3 years ago

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Write the number in word form.

Evgesh-ka [11]

Answer:

sixty-six millionths
0.027
$12.60

Step-by-step explanation:

(a) The least-significant digit is in the 6th place to the right of the decimal. Numbers in that place have the value "millionths." The number is 66 times 1 millionth, or sixty-six millionths.

(b) 27/1000 = 0.027 . . . . the least-significant digit goes in the "thousandths" place

(c) The digit to the right of the "cents" place is 5. This is more than 4, so 1 must be added to the cents place when rounding. $12.59 + 0.01 = $12.60.

8 0

3 years ago

Please help me with the below question.

Alexus [3.1K]

a) Substitute $y=x^9$ and $dy=9x^8\,dx$ :

$\displaystyle \int x^8 \cos(x^9) \, dx = \frac19 \int 9x^8 \cos(x^9) \, dx \\\\ = \frac19 \int \cos(y) \, dy \\\\ = \frac19 \sin(y) + C \\\\ = \boxed{\frac19 \sin(x^9) + C}$

b) Integrate by parts:

$\displaystyle \int u\,dv = uv - \int v \, du$

Take $u = \ln(x)$ and $dv=\frac{dx}{x^7}$ , so that $du=\frac{dx}x$ and $v=-\frac1{6x^6}$ :

$\displaystyle \int \frac{\ln(x)}{x^7} \, dx = -\frac{\ln(x)}{6x^6} + \frac16 \int \frac{dx}{x^7} \\\\ = -\frac{\ln(x)}{6x^6} + \frac1{36x^6} + C \\\\ = \boxed{-\frac{6\ln(x) + 1}{36x^6} + C}$

c) Substitute $y=\sqrt{x+1}$ , so that $x = y^2-1$ and $dx=2y\,dy$ :

$\displaystyle \frac12 \int e^{\sqrt{x+1}} \, dx = \frac12 \int 2y e^y \, dy = \int y e^y \, dy$

Integrate by parts with $u=y$ and $dv=e^y\,dy$ , so $du=dy$ and $v=e^y$ :

$\displaystyle \int ye^y \, dy = ye^y - \int e^y \, dy = ye^y - e^y + C = (y-1)e^y + C$

Then

$\displaystyle \frac12 \int e^{\sqrt{x+1}} \, dx = \boxed{\left(\sqrt{x+1}-1\right) e^{\sqrt{x+1}} + C}$

d) Integrate by parts with $u=\sin(\pi x)$ and $dv=e^x\,dx$ , so $du=\pi\cos(\pi x)\,dx$ and $v=e^x$ :

$\displaystyle \int \sin(\pi x) \, e^x \, dx = \sin(\pi x) \, e^x - \pi \int \cos(\pi x) \, e^x \, dx$

By the fundamental theorem of calculus,

$\displaystyle \int_0^1 \sin(\pi x) \, e^x \, dx = - \pi \int_0^1 \cos(\pi x) \, e^x \, dx$

Integrate by parts again, this time with $u=\cos(\pi x)$ and $dv=e^x\,dx$ , so $du=-\pi\sin(\pi x)\,dx$ and $v=e^x$ :

$\displaystyle \int \cos(\pi x) \, e^x \, dx = \cos(\pi x) \, e^x + \pi \int \sin(\pi x) \, e^x \, dx$

By the FTC,

$\displaystyle \int_0^1 \cos(\pi x) \, e^x \, dx = e\cos(\pi) - 1 + \pi \int_0^1 \sin(\pi x) \, e^x \, dx$

Then

$\displaystyle \int_0^1 \sin(\pi x) \, e^x \, dx = -\pi \left(-e - 1 + \pi \int_0^1 \sin(\pi x) \, e^x \, dx\right) \\\\ \implies (1+\pi^2) \int_0^1 \sin(\pi x) \, e^x \, dx = 1 + e \\\\ \implies \int_0^1 \sin(\pi x) \, e^x \, dx = \boxed{\frac{\pi (1+e)}{1 + \pi^2}}$

e) Expand the integrand as

$\dfrac{x^2}{x+1} = \dfrac{(x^2 + 2x + 1) - (2x+1)}{x+1} = \dfrac{(x+1)^2 - 2 (x+1) + 1}{x+1} \\\\ = x - 1 + \dfrac1{x+1}$

Then by the FTC,

$\displaystyle \int_0^1 \frac{x^2}{x+1} \, dx = \int_0^1 \left(x - 1 + \frac1{x+1}\right) \, dx \\\\ = \left(\frac{x^2}2 - x + \ln|x+1|\right)\bigg|_0^1 \\\\ = \left(\frac12-1+\ln(2)\right) - (0-0+\ln(1)) = \boxed{\ln(2) - \frac12}$

f) Substitute $e^{7x} = \tan(y)$ , so $7e^{7x} \, dx = \sec^2(y) \, dy$ :

$\displaystyle \int \frac{e^{7x}}{e^{14x} + 1} \, dx = \frac17 \int \frac{\sec^2(y)}{\tan^2(y) + 1} \, dy \\\\ = \frac17 \int \frac{\sec^2(y)}{\sec^2(y)} \, dy \\\\ = \frac17 \int dy \\\\ = \frac y7 + C \\\\ = \boxed{\frac17 \tan^{-1}\left(e^{7x}\right) + C}$

8 0

2 years ago

4(r+3)-3r=19 help me plz

bulgar [2K]

Since ur using r as a variable the answer is r = 7.

6 0

3 years ago

Read 2 more answers