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jonny [76]
2 years ago
11

Pleasee help!!!! 20 points

Mathematics
1 answer:
Usimov [2.4K]2 years ago
3 0

The function with the same end behavior of f(x) is the last option:

g(x) = (-\frac{1}{3} )*x

<h3>Which function has the same end behavior of f(x)?</h3>

By looking at the graph, we can see that the end behavior of f(x) is:

as x → ∞; f(x) → -∞

as x → -∞; f(x) → ∞

Then we just want a function that depends linearly with x and that has a negative coefficient.

The correct option is the last one:

g(x) = (-\frac{1}{3} )*x

Where clearly, when x tends to infinity the function tends to negative infinity, and when x tends to negative infinity, the function tends to infinity.

If you want to learn more about end behavior:

brainly.com/question/1365136

#SPJ1

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Simplify . (1/c + 1/h)/(1/(c ^ 2) - 1/(r ^ 2))
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Answer:

\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}

Step-by-step explanation:

\frac{\frac{1}{c}+\frac{1}{h}}{\frac{1}{c^2}-\frac{1}{r^2}}

Combine \frac{1}{c} + \frac{1}{h}

\frac{\frac{h+c}{ch}}{\frac{1}{c^2}-\frac{1}{r^2}}

Combine the bottom, too.

=\frac{\frac{h+c}{ch}}{\frac{r^2-c^2}{c^2r^2}}

Apply the fraction rule

=\frac{\left(h+c\right)c^2r^2}{ch\left(r^2-c^2\right)}

Cancel

=\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}

Therefore, \frac{\left(\frac{1}{c}+\frac{1}{h}\right)}{\left(\frac{1}{\left(c^2\right)}-\frac{1}{\left(r^2\right)}\right)}:\quad \frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}

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3 years ago
What is equivalent to 5-squared3
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Answer: 125

Step-by-step explanation:

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A projector was sold after allowing 10% discount on the marked price and levying 13% VAT .If the selling price of the projector
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The marked price of the projector is $50,000.

<h3>What is the marked price?</h3>

The price of the projector after the discount can be represented with:

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The price of the projector after the VAT is : (1.13) x 0.90x = 1.02x

Difference in price = 1.02x - 0.90x = 5850

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x = 5850 / 0.12

x = $50,000

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