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3 years ago

15 3/4% is equal to which decimal? A- 0.1575 B- 157.25 C-15.25 D- 15.34

Mathematics

2 answers:

eimsori [14]3 years ago

6 0

15 3/4% = 15.75%
15.75% = 0.1575
The answer is A.

stepan [7]3 years ago

6 0

The only one that would make sense is: A.

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The city's central library has 72,864 books. The library has 72 bookcases. If each bookcase has room for the same number of book

solong [7]

Answer:

There are 1012 books per bookcase

Step-by-step explanation:

To find the number of books per bookcase take the number of bookcases (72) divided by the number of books (72,864)

So;

72,864 / 72 = 1012

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3 years ago

Read 2 more answers

Select the correct answer.

Setler79 [48]

Answer:

A. 2·x² + 16·x + 32 ≥ 254

Step-by-step explanation:

The given dimensional relationship between the dimensions of the photo in the center of the cake and the dimensions of the cake are

The width of the cake = The width of the photo at the center of the cake, x + 4 inches

The length of the cake = 2 × The width of the cake

The area of the cake Wanda is working on ≥ 254 in.²

Where 'x' represents the width of the photo (at the center of the cake), let 'W' represent the width of the cake, let 'L' represent the length of the cake, we get;

W = x + 4

L = 2 × W

Area of the cake, A = W × L ≥ 254

∴ A = (x + 4) × 2 × (x + 4) = 2·x² + 16·x + 32 ≥ 254

The inequality representing the solution is therefore;

2·x² + 16·x + 32 ≥ 254

3 0

3 years ago

PLEASE HELP!! WHICH ONE IS GREATER?? -3 or 3 PLEASE EXPLANE IF YOU CAN??! THANK YOU!

jarptica [38.1K]

3 because Negative numbers equal less than 0. See the number line attached below. In the number line, all the negative numbers are on the left of the zero and all the positive numbers are on the right of the 0.

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4 years ago

Given f: ℝ → ℝ and : ℝ → ℝ , for the following, find f(g(x)) and g(f(x)), and state the domain.

horrorfan [7]

Answer:

Remember, $(f\circ g)(x)=f(g(x))$ and the range of g must be in the domain of f.

$f(g(x))=f(x-1)=(x-1)^2-(x-1)=x^2-2x+1-x+1=x^2-3x+2$

$g(f(x))=g(x^2-x)=(x^2-x)-1=x^2-x-1$

The domain of f(g(x)) and g(f(x)) is the set of reals.

$f(g(x))=f(\sqrt{x}-2)=(\sqrt{x}-2)^2-x=\sqrt{x}^2-2*2*\sqrt{x}+2^2-x=-4\sqrt{x}+4$

$g(f(x))=g(x^2-x)=\sqrt{x^2-x}-2$

The domain of f(g(x)) is the set of nonnegative reals and the domain of g(f(x)) is the set of number such that $x^2-x\geq 0$

$f(g(x))=f(\frac{1}{x-1})=(\frac{1}{x-1})^2=\frac{1}{(x-1)^2}$

$g(f(x))=g(x^2)=\frac{1}{x^2-1}$

The domain of f(g(x)) is the set of reals except the 1 and the domain of g(f(x)) is the set of reals except the 1 and -1

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The domain of f(g(x)) is the set of reals except 2, and the domain of g(f(x)) is the set of reals except -1.

$f(g(x))=f(log(2(x+3)))=f(log(2x+6))=log(2x+6)-1$

$g(f(x))=g(x-1)=log(2(x-1)+6)=log(2x+4)$

The domain of f(g(x)) is the set of nonnegative reals except -3. The domain of g(f(x)) is the set of nonnegative reals except -2.

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3 years ago