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8090 [49]
3 years ago
15

If f(x) = x4 − x3 + x2 and g(x) = −x2, where x ≠ 0, what is (f ⁄g)(x)?

Mathematics
2 answers:
VikaD [51]3 years ago
7 0
I posted an image instead.

Nina [5.8K]3 years ago
7 0

C)  -x^2 + x - 1

Is the answer

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A large pot holds 12 gallons of soup Jared has 1-pint containers of chicken broth. Complete the table to help you find the numbe
Rina8888 [55]
There is 8 pints in 1 gallon so 12x8=96
96 pints to fill a 12 gallon pot
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3 years ago
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If f(x) = 3a|4x - 4| - ax, where a is some constant, find f '(1). (4 points)
mihalych1998 [28]

Annswer girl

Step-by-step explanation:

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8 0
4 years ago
Use the Remainder Theorem to determine which binomial is a factor of the function f(x) = x3 − 2x2 − 19x + 20.
dybincka [34]

Answer:

(x - 1)

Step-by-step explanation:

Given

f(x) = x³ - 2x² - 19x + 20

The sum of the coefficients is 1 - 2 - 19 + 20 = 0

This means that x = 1 is a root and (x - 1) is a factor of f(x)

3 0
3 years ago
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1. Whats the exact value of Y?<br> 2. Whats the exact value of X?
Alexxx [7]

Answer:

Hello! The answer is in this link.

Step-by-step explanation:

8518-06-04-10-15-instructional.mp4. I hope this helps! Have a great rest of your day!

4 0
3 years ago
If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,
julsineya [31]
<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

<h2>Why?</h2>

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)

(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

5 0
4 years ago
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