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3 years ago

What decimal is equivalent to 500 thousandths and 8 hundredths?

2 answers:

7 0

500 thousandths is equal to five tenths. Thus, we have a five in the tenths place and an eight in the hundredths place, so we have .58 as our answer.

Stels [109]3 years ago

6 0

.58 is your answer hope it helps your welcome

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HELP ASAP WILL MARK BRAINLIEST!<br><br> What is the length of TW?

love history [14]

Answer:

Step-by-step explanation:

You are given the hypotenuse and one leg of a right triangle, and are being asked to find the other leg. The Pythagorean theorem applies.

TW² + TY² = WY²

TW² = 17² - 15² = 64

TW = √64

TW = 8

3 0

3 years ago

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Which of the following is equivalent to 60 Superscript one-half?

vitfil [10]

Answer:c

Step-by-step explanation:

The Answer is C.

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4 years ago

I need help plzzzzzzzzzzzzzzzzzzzzz

blagie [28]

Answer:

Step-by-step explanation:

Start box

180 = 89+42+x

180-89-42=x

49 = x

2nd box

180 = 84+58+x

180-84-58=x

38 = x

3rd box

180=74+2x

180-74=2x

106/2 = x

53 = x

4th box

180=102+2x

180-102=2x

78/2 = x

39 = x

4th box

180 = 73+81 +x

180-73-81=x

26 = x

5th box

180=2*54+x

180 - 2*54 = x

72 = x

6th box

180=62-2x

180-62=2x

118=2x

118/2=x

59 = x

7th box

180 = 2*68 + x

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3 0

3 years ago

A group of students is collecting books to add to their librarby. The table shows the number of books in the librarby after 1, 3

Grace [21]

Answer:

(7,48)

Step-by-step explanation

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6 0

3 years ago

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What is the antiderivative of 3x/((x-1)^2)

Maslowich

Answer:

$\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

Step-by-step explanation:

Given

$\int \:\:3\cdot \frac{x}{\left(x-1\right)^2}dx$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=3\cdot \int \frac{x}{\left(x-1\right)^2}dx$

$\mathrm{Apply\:u-substitution:}\:u=x-1$

$=3\cdot \int \frac{u+1}{u^2}du$

$\mathrm{Expand}\:\frac{u+1}{u^2}:\quad \frac{1}{u}+\frac{1}{u^2}$

$=3\cdot \int \frac{1}{u}+\frac{1}{u^2}du$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$=3\left(\int \frac{1}{u}du+\int \frac{1}{u^2}du\right)$

$\int \frac{1}{u}du=\ln \left|u\right|$ ∵ $\mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)$

$\int \frac{1}{u^2}du=-\frac{1}{u}$ ∵ $\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1$

$=3\left(\ln \left|u\right|-\frac{1}{u}\right)$

$\mathrm{Substitute\:back}\:u=x-1$

$=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)$

$\mathrm{Add\:a\:constant\:to\:the\:solution}$

$=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

Therefore,

$\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

4 0

3 years ago