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2 years ago

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Tomas learned that the product of the polynomials (a + b)(a2 – ab + b2) was a special pattern that would result in a sum of cube

s, a3 + b3. His teacher put four products on the board and asked the class to identify which product would result in a sum of cubes if a = 2x and b = 1.
Which product should Tomas choose?

A) (2x + 1)(2x2 + 2x – 1)
B) (2x + 1)(4x3 + 2x – 1)
C) (2x + 1)(4x2 – 2x + 1)
D) (2x + 1)(2x2 – 2x + 1)

2 answers:

9966 [12]2 years ago

7 0

(a+b)(a²-ab+b²)
(2x+1)((2x)²-(2x)(1)+1²)
(2x+1)(4x²-2x+1)

C is answer

Anna [14]2 years ago

3 0

General form: a^3 + b^3 = (a + b) (a^2 -ab +b^2)

Now replace with a= 2x, b = 1

(2x)^3 + 1^3 = (2x + 1) [ (2x)^2 -(2x)(1) + 1^2) = (2x + 1) (4x^2 - 2x + 1)

Therefore, the answer is the option C)

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The principal amount was $23,393.45

Step-by-step explanation:

The total amount paid on a 35 year loan was $98,000 at the rate of interest 4.1%

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2 years ago

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Arrange these functions from the greatest to the least value based on the average rate of change in the specified interval.Tiles

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By definition, the average rate of change is given by:

$AVR = \frac{f(x2)-f(x1)}{x2-x1}$

We evaluate each of the functions in the given interval.

We have then:

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Evaluating for x = -2:

$f (-2) = (-2) ^ 2 + 3 (-2)\\f (-2) = 4 - 6\\f (-2) = - 2$

Evaluating for x = 3:

$f (3) = (3) ^ 2 + 3 (3)\\f (3) = 9 + 9\\f (3) = 18$

Then, the AVR is:

$AVR = \frac{18-(-2)}{3-(-2)}$

$AVR = \frac{18+2}{3+2}$

$AVR = \frac{20}{5}$

$AVR = 4$

For f (x) = 3x - 8:

Evaluating for x =4:

$f (4) = 3 (4) - 8\\f (4) = 12 - 8\\f (4) = 4$

Evaluating for x = 5:

$f (5) = 3 (5) - 8\\f (5) = 15 - 8\\f (5) = 7$

Then, the AVR is:

$AVR = \frac{7-4}{5-4}$

$AVR = \frac{3}{1}$

$AVR = 3$

For f (x) = x ^ 2 - 2x:

Evaluating for x = -3:

$f (-3) = (-3) ^ 2 - 2 (-3)\\f (-3) = 9 + 6\\f (-3) = 15$

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$f (4) = (4) ^ 2 - 2 (4)\\f (4) = 16 - 8\\f (4) = 8$

Then, the AVR is:

$AVR = \frac{8-15}{4-(-3)}$

$AVR = \frac{-7}{4+3}$

$AVR = \frac{-7}{7}$

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For f (x) = x ^ 2 - 5:

Evaluating for x = -1:

$f (-1) = (-1) ^ 2 - 5\\f (-1) = 1 - 5\\f (-1) = - 4$

Evaluating for x = 1:

$f (1) = (1) ^ 2 - 5\\f (1) = 1 - 5\\f (1) = - 4$

Then, the AVR is:

$AVR = \frac{-4-(-4)}{1-(-1)}$

$AVR = \frac{-4+4}{1+1}$

$AVR = \frac{0}{2}$

$AVR = 0$

Answer:

from the greatest to the least value based on the average rate of change in the specified interval:

f(x) = x^2 + 3x interval: [-2, 3]

f(x) = 3x - 8 interval: [4, 5]

f(x) = x^2 - 5 interval: [-1, 1]

f(x) = x^2 - 2x interval: [-3, 4]

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Answer:

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