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3 years ago

Let f(x) = 4x3 + 3x2 − 20x − 15 and g(x) = 4x + 3. Find f of x over g of x.

Mathematics

1 answer:

serg [7]3 years ago

8 0

Answer:

$4x^2+15x+15+\frac{35}{x-3}$

Step-by-step explanation:

Let f(x) = 4x3 + 3x2 − 20x − 15 and g(x) = 4x + 3. Find f of x over g of x.

$\frac{f(x)}{g(x)} \\\frac{4x^3 + 3x^2 - 20x - 15}{4x+3}$

$4x^2+15x+15+\frac{35}{x-3}$

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What are the roots of this quartaic equation? -102 + 12r - 9 = 0

elena-14-01-66 [18.8K]

Answer:

$r = \frac{37}{4}$

I hope this helps!

7 0

2 years ago

Read 2 more answers

In a school election, Juan received 4 times as many votes as Wayne, Neal recurved twenty less votes than Juan, and Kerry got hal

77julia77 [94]

Votes received by Wayne is 112

Solution:

To find: votes received by Wayne

Let the vote received by Wayne be "x"

Juan received 4 times as many votes as Wayne

Therefore,

Juan votes = 4 times as many votes as Wayne

Juan votes = 4x ---- eqn 1

Neal received twenty less votes than Juan

Neal votes = twenty less votes than Juan

Neal votes = Juan votes - 20

Neal votes = 4x - 20 ---- eqn 2

Kerry got half as many votes as Neal

Kerry votes = half of neal votes

Kerry votes = $\frac{4x - 20}{2}$ ---- eqn 3

The total votes cast in the election was 1,202

Wayne votes + Juan votes + Neal votes + Kerry votes = 1202

Plug in eqn 1 , eqn 2, eqn 3

$x + 4x + 4x - 20 + \frac{(4x - 20)}{2} = 1202\\\\2x + 8x + 8x - 40 + 4x - 20 = 1202 \times 2\\\\22x - 60 = 2404\\\\22x = 2404 + 60\\\\22x = 2464\\\\x = 112$

Therefore votes received by Wayne is 112

4 0

3 years ago

When 31 times a number is increased by 40 the answer is the same as two 200 is decreased by the number

storchak [24]

$x-\ number\\\\
31x+40=200-x\ \ \ | add\ x\\\\
32x+40=200\ \ \ | subtract\ 40\\\\
32x=160\ \ \ \ | divide\ by\ 32\\\\x=5\\\\Number\ is\ 5.$

4 0

3 years ago

A professor pays 25 cents for each blackboard error made in lecture to the student who pointsout the error. In a career ofnyears

marta [7]

Answer:

(a) The probability that Y₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b) n = 28.09

Step-by-step explanation:

The random variable Yₙ is defined as the total numbers of dollars paid in n years.

It is provided that Yₙ can be approximated by a Gaussian distribution, also known as Normal distribution.

The mean and standard deviation of Yₙ are:

$\mu_{Y_{n}}=40n\\\sigma_{Y_{n}}=\sqrt{100n}$

(a)

For n = 20 the mean and standard deviation of Y₂₀ are:

$\mu_{Y_{n}}=40n=40\times20=800\\\sigma_{Y_{n}}=\sqrt{100n}=\sqrt{100\times20}=44.72\\$

Compute the probability that Y₂₀ exceeds 1000 as follows:

$P(Y_{n}>1000)=P(\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}>\frac{1000-800}{44.72})\\=P(Z> 4.47)\\=1-P(Z$

**Use a z table for probability.

Thus, the probability that Y₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b)

It is provided that P (Yₙ > 1000) > 0.99.

$P(Y_{n}>1000)=0.99\\1-P(Y_{n}$

The value of z for which P (Z < z) = 0.01 is 2.33.

Compute the value of n as follows:

$z=\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}\\2.33=\frac{1000-40n}{\sqrt{100n}}\\2.33=\frac{100}{\sqrt{n}}-4\sqrt{n} \\2.33=\frac{100-4n}{\sqrt{n}} \\5.4289=\frac{(100-4n)^{2}}{n}\\5.4289=\frac{10000+16n^{2}-800n}{n}\\5.4289n=10000+16n^{2}-800n\\16n^{2}-805.4289n+10000=0$

The last equation is a quadratic equation.

The roots of a quadratic equation are:

$n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

a = 16

b = -805.4289

c = 10000

On solving the last equation the value of n = 28.09.

8 0

3 years ago

Simplify the following expression: NEED HELP QUICK PLZ :(

Anna71 [15]

The answer is B

-8y + 7y = -y

When the signs are different, subtract and take the sign of the bigger number.

5 0

2 years ago