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3 years ago

2) Find the remainder when x3 + x2 - 20 is divided by

Mathematics

1 answer:

Dahasolnce [82]3 years ago

4 0

Answer:

Remainder = –24

Step-by-step explanation:

x³ + x² – 20 divide by x + 2

We can obtain the remainder when

x³ + x² – 20 is divided by x + 2 using the remainder theorem as follow:

We'll begin by calculating the value of x from x + 2. This is illustrated below:

x + 2 = 0

Collect like terms

x = 0 – 2

x = –2

Finally, we shall determine the remainder by substituting –2 for x in

x³ + x² – 20. This is illustrated below:

x³ + x² – 20

x = –2

Remainder = (–2)³ + (–2)² – 20

Remainder = –8 + 4 – 20

Remainder = –24

Therefore, the ramainder when

x³ + x² – 20 is divided by x + 2 is –24

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4 years ago

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Answer: The value of other side is 7.8 inch.

Explanation:

It is given that the isosceles triangle has angle measures 50°, 50°, and 80°. The side across from the 80° angle is 10 inches long.

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$\frac{a}{\sin A}= \frac{b}{\sin B} =\frac{c}{\sin C}$

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4 years ago

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zhannawk [14.2K]

We conclude that the profit if the company produces 10 units is 1,040 dollars.

<h3></h3><h3>How to find the profit?</h3>

The profit is defined as the difference between the revenue and the costs.

In this case, we know that the revenue and costs are defined by:

R(x) = 200x - 5x^2

C(x) = 360 + 6x

Then the profit equation is:

P(x) = R(x) - C(x).

P(x) = 200x - 5x^2 - (360 + 6x)

P(x) = -5x^2 + 194x - 360

Now we want to find the profit when x = 10, then we will get:

p(10) = -5*10^2 + 194*10 - 360 = 1,040

We conclude that the profit if the company produces 10 units is 1,040 dollars.

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2 years ago

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Arte-miy333 [17]

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3 years ago

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In a box of 10 calculators, one is defective. In how many ways can four calculators be selected, if you know one

Juliette [100K]

Given:

Total number of calculators in a box = 10

Defective calculators in the box = 1

To find:

The number of ways in which four calculators be selected and one of the four calculator is defective.

Solution:

We have,

Total calculators = 10

Defective calculators = 1

Then, Non-defective calculator = 10-1 = 9

Out of 4 selected calculators 1 should be defective. So, 3 calculators are selected from 9 non-defective calculators and 1 is selected from the defective calculator.

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$\text{Total ways}=\dfrac{9!}{3!(9-3)!}\times 1$

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Therefore, the four calculators can be selected in 84 ways.

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3 years ago