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Alchen [17]
2 years ago
7

The polynomial p(x)=x^3-6x^2+32p(x)=x 3 −6x 2 +32p, left parenthesis, x, right parenthesis, equals, x, cubed, minus, 6, x, squar

ed, plus, 32 has a known factor of (x-4)(x−4)left parenthesis, x, minus, 4, right parenthesis. Rewrite p(x)p(x)p, left parenthesis, x, right parenthesis as a product of linear factors.
Mathematics
1 answer:
Ray Of Light [21]2 years ago
8 0

Answer:

(x-4)(x-4)(x+2)

Step-by-step explanation:

Given p(x) = x^3-6x^2+32 when it is divided by  x - 4, the quotient gives

x^2-2x-8

Q(x) = P(x)/d(x)

x^3-6x^2+32/x- 4  =  x^2-2x-8

Factorizing the quotient

x^2-2x-8

x^2-4x+2x-8

x(x-4)+2(x-4)

(x-4)(x+2)

Hence the polynomial as a product if linear terms is (x-4)(x-4)(x+2)

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What number will the function return if the input is 15.4?
wlad13 [49]

The rule of the function is to multiply the input by 3, since one yard is equal in length to three feet.

So, if the input is 15.4, the output will be

f(x)=3x \implies f(15.4)=3\cdot 15.4 = 46.2

4 0
3 years ago
Caleb borrowed $1500 from his aunt. He plans to pay his aunt back in 9 months. If he
sergeinik [125]

Answer:

A = $1,545.00

(I = A - P = $45.00)

Equation:

A = P(1 + rt)

Explanation:

First, converting R percent to r a decimal

r = R/100 = 4%/100 = 0.04 per year.

Putting time into years for simplicity,

9 months / 12 months/year = 0.75 years.

Solving our equation:

A = 1500(1 + (0.04 × 0.75)) = 1545

A = $1,545.00

The total amount accrued, principal plus interest, from simple interest on a principal of $1,500.00 at a rate of 4% per year for 0.75 years (9 months) is $1,545.00.

4 0
3 years ago
Susan needs to buy 3 helium balloons. She can buy individual balloons for $2.75 each, or a 3-pack of balloons for $7.80. How muc
Mrac [35]

Answer:

15

Step-by-step explanation:

8.25 minus 7.80 is 0.45 45 divided by 3 is 15 so the answer is 15

8 0
3 years ago
Given: PRST is a square
xxTIMURxx [149]

Answer:

(1-\sqrt{2})a^2

Step-by-step explanation:

Consider irght triangle PRS. By the Pythagorean theorem,

PS^2=PR^2+RS^2\\ \\PS^2=a^2+a^2\\ \\PS^2=2a^2\\ \\PS=\sqrt{2}a

Thus,

MS=PS-PM=\sqrt{2}a-a=(\sqrt{2}-1)a

Consider isosceles triangle MSC. In this triangle

MS=MC=(\sqrt{2}-1)a.

The area of this triangle is

A_{MSC}=\dfrac{1}{2}MS\cdot MC=\dfrac{1}{2}\cdot (\sqrt{2}-1)a\cdot (\sqrt{2}-1)a=\dfrac{(\sqrt{2}-1)^2a^2}{2}=\dfrac{(3-2\sqrt{2})a^2}{2}

Consider right triangle PTS. The area of this triangle is

A_{PTS}=\dfrac{1}{2}PT\cdot TS=\dfrac{1}{2}a\cdot a=\dfrac{a^2}{2}

The area of the quadrilateral PMCT is the difference in area of triangles PTS and MSC:

A_{PMCT}=\dfrac{(3-2\sqrt{2})a^2}{2}-\dfrac{a^2}{2}=\dfrac{(2-2\sqrt{2})a^2}{2}=(1-\sqrt{2})a^2

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3 years ago
Prove that the two circles shown are similar
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Zoom factor will be the number to prove they’re the same
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