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Daniel [21]
2 years ago
5

Select all polynomials that are divisible by (x-1)(x−1)left parenthesis, x, minus, 1, right parenthesis. Choose all answers that

apply: Choose all answers that apply: (Choice A) A A(x)=3x^3+2x^2-xA(x)=3x 3 +2x 2 −xA, left parenthesis, x, right parenthesis, equals, 3, x, cubed, plus, 2, x, squared, minus, x (Choice B) B B(x)=5x^3-4x^2-xB(x)=5x 3 −4x 2 −xB, left parenthesis, x, right parenthesis, equals, 5, x, cubed, minus, 4, x, squared, minus, x (Choice C) C C(x)=2x^3-3x^2+2x-1C(x)=2x 3 −3x 2 +2x−1C, left parenthesis, x, right parenthesis, equals, 2, x, cubed, minus, 3, x, squared, plus, 2, x, minus, 1 (Choice D) D D(x)=x^3+2x^2+3x+2D(x)=x 3 +2x 2 +3x+2
Mathematics
1 answer:
alexdok [17]2 years ago
8 0

Answer:

Step-by-step explanation:

For us to be able to determine the polynomials that are divisible by (x-1), this means that x-1 must be a factor for the functon to be able to divide any of the polynimial.

Since x-1 is a factor, we can get the value of x

x-1 = 0

x =0+1

x = 1

Next is for to substitute x - 1 into the polynomial and see the ones that will give us zero

For A(x)=3x^3+2x^2-x

A(1) = 3(1)^3+2(1)^2-(1)

A(1) = 3+2-(1)

A(1) = 5-1

A(1) = 4

Since A(1) ≠ 0, then x-1 is not divisible by the polynomial function.

<u>For B(x)=5x^3-4x^2-x</u>

B(1)=5(1)^3-4(1)^2-1

B(1)=5-4-1

B(1)=1-1 = 0

Since B(1) = 0, hence x-1 is divisible by 5x^3-4x^2-x

For the polynomial  C(x)= 2x^3-3x^2+2x-1

C(1)=2(1)^3-3(1)^2+2(1)-1

C(1)=2-3+2-1

C(1)= -1+1

C(1)= 0

Since C(1) = 0, hence x-1 is divisible by<u> the </u>

<u />

<u>F</u>or the polynomial D(x)=x^3+2x^2+3x+2

D(1)=1^3+2(1)^2+3(1)+2

D(1)=1+2+3+2

D(x) = 8

Hence the polynomial D(x) is not divisible by x-1

Hence the correct options are B(x)=5x^3-4x^2-x and 2x^3-3x^2+2x-1

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4 0
2 years ago
Solve the given initial value problem and determine how the interval in which the solution exists depends on the initial value y
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Answer:

y has a finite solution for any value y_0 ≠ 0.

Step-by-step explanation:

Given the differential equation

y' + y³ = 0

We can rewrite this as

dy/dx + y³ = 0

Multiplying through by dx

dy + y³dx = 0

Divide through by y³, we have

dy/y³ + dx = 0

dy/y³ = -dx

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-1/(2y²) = - x + c

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1/(2y²) = x + C (Where C = -c)

Applying the initial condition y(0) = y_0, put x = 0, and y = y_0

1/(2y_0²) = 0 + C

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2 years ago
The inequality that will determine the number of months, x, that are required for the second phone to be less expensive is . The
ziro4ka [17]

<em>Question:</em>

<em>Sal is trying to determine which cell phone and service plan to buy for his mother. The first phone costs $100 and $55 per month for unlimited usage. The second phone costs $150 and $51 per month for unlimited usage. </em>

<em>The inequality that will determine the number of months, x, that are required for the second phone to be less expensive is . </em>

<em>The solution to the inequality is . </em>

<em>Sal’s mother would have to keep the second cell phone plan for at least months in order for it to be less expensive.</em>

Answer:

a. 150 + 51x < 100 + 55x

b. x > 12.5

c. At least 13 months

Step-by-step explanation:

Given

First Phone;

Cost = \$100

Additional = \$55 <em>(monthly)</em>

Second Phone;

Cost = \$150

<em />Additional = \$51<em> (monthly)</em>

<em></em>

Solving (a): The inequality

<em></em>

Represent the number of months with x

The first phone is expressed as:

100 + 55x

The second phone is expressed as:

150 + 51x

For the second to be less expensive that the first, the inequality is:

150 + 51x < 100 + 55x

Solving (b): Inequality Solution

150 + 51x < 100 + 55x

Collect Like Terms

51x-55x

-4x

Solve for x

x > -50/-4

x > 12.5

Solving (c): Interpret the solution in (b)

x > 12.5 implies 13, 14, 15....

Hence, She'll keep the second phone for a period of at least 13 months

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