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3 years ago

Can anyone help me answer this

Mathematics

1 answer:

frozen [14]3 years ago

5 0

Answer:

TV 46 115 138 161 253

Defects 2 5 6 7 11

Step-by-step explanation: I got that rate of change Times 23, so i plug it in to T / 23, D / 23, T = TV's, D = Defects

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I need help plz <br> thanks

m_a_m_a [10]

What are the answer choices in the drop down?

4 0

2 years ago

Y + 2/5x =1 what does y equal?

bezimeni [28]

The answer is Y=-2/5x+1 :)

5 0

3 years ago

Read 2 more answers

The average weight of three dogs is 55 pounds, and the average weight of five cats is 9 pounds. What is the average weight, in p

notsponge [240]

Answer:

Option D is correct

Step-by-step explanation:

Statement 1

Average weight of 3 dogs = 55 pounds

Total weight of 3 dogs = 55 × 3

Total weight of 3 dogs = 165 pound

Statement 2

Average weight of 5 cats = 9 pounds

Total weight of 5 cats = 9 × 5

Total weight of 5 cats = 45 pounds

New Average = 165 + 45/8

New Average = 210/8

New Average = 26.25 pounds

4 0

3 years ago

Read 2 more answers

7+3√5/3+√5 - 7-3√5/3-√5 = a + b√5

daser333 [38]

<h3>Given:-</h3>

$\\ \sf \implies\frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } = a + \sqrt{5} b \\$

The value of a and b

<h3>Solution:-</h3>

$\\ \sf \implies\frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } = a + \sqrt{5} b \\$

$\\ \sf \implies\frac{( \: 7 + 3 \sqrt{5} \: \: ( 3 - \sqrt{5}) \: \: - 7 - 3 \sqrt{5} \: \: ( 3 + \sqrt{5}) \: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{( \: 21 - 7 \sqrt{5} \: + 9 \sqrt{5} - 15) \: \: - ( \: 21 + 7 \sqrt{5} \: - 9 \sqrt{5} + 15)\: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{( \: 6 + 2 \sqrt{5} ) \: \: - ( \: 6 - 2 \sqrt{5} )\: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: 6 + 2 \sqrt{5} \: \: - \: \: 6 - 2 \sqrt{5} \: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: 4 \sqrt{5} \: \: }{3 {}^{2} - {\sqrt{5} }^{2} } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: 4 \sqrt{5} \: \: }{ \: \: \: \: 9 - 5 \: \: \: } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: 4 \sqrt{5} \: \: }{ \: \: \: \: 4 \: \: \: } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: \cancel{4 } \sqrt{5} \: \: }{ \: \: \: \: \cancel{4 }\: \: \: } = a + \sqrt{5} \: b\\$

$\\ \sf \implies \: \sqrt{5} = a + \sqrt{5} \: b\\$

we can also write it as ;

$\\ \sf \implies \: 0 + \sqrt{5} = a + \sqrt{5} \: b\\$

★<u> </u><u>Henceforth, the value of a and b are</u> :

→ a = 0

→ b = 1

6 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

3 years ago