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3 years ago

Use the diagram to fi d the missing lenght of the pool (please i need the help)

Mathematics

1 answer:

zlopas [31]3 years ago

6 0

Answer:

D) 50ft

Step-by-step explanation:

(18×x) + (32×18) + [(48-18-18)×(32-11)]

= 1728

18x + 576 + (12×21) = 1728

18x = 1728 - 828

18x = 900

x = 50

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ch4aika [34]

The radius is half the diameter. So the diameter is 14.

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2 years ago

3x - 1 = -5?

Setler79 [48]

Answer:

3x - 1 = -5?

What is x if

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-20

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Step-by-step explanation:

8 0

3 years ago

Let f(t) ANSWER: 1+ = t² - 1 t+3 Find f'(1)

Luda [366]

Answer:

f(1) = 3

Step-by-step explanation:

f(t) = t² - 1t + 3

we want to find f(1)

all we have to do is replace the t's with 1 and evaluate

f(1) = (1)² - 1(1) + 3

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4 0

1 year ago

The directions for a homemade glass cleaner call for 1 part vinegar to 4 parts water. Mr. Gilbert mixed 2 cups of vinegar with 5

PolarNik [594]

No.
EXPLANATION:
To tackled this problem we should find an equivalent denominator.

We find that the lcm ( lowest common denominator ) is 20.

1 x 5 = 5
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2 x 4 = 8
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so the answer is no.

3 0

2 years ago

Read 2 more answers

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

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

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

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.

3 0

2 years ago