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lina2011 [118]
3 years ago
5

Does someone know the answers and how to solve them?

Mathematics
1 answer:
il63 [147K]3 years ago
4 0

Answer: D

Step-by-step explanation:

Yeah, lets look at each of the functions. I'll refer to them as A, B, C, D, and E.

A) f(x)=x^2-16

This has two zero. x^2-16= (x+4)(x-4), which two zeroes include -4 and 4.

Here we introduce an important concept, if the function isn't a perfect square, and is a quadratic, it has two zeroes.

B) This is linear, so it obviously has 1 zero.

C) Two zeroes, because it isn't a perfect square.

D) x^2-10x+25 = (x-5)^2, so it has only one zero.

E) Not a perfect square, so two roots.

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Figure ABC is to be translated to Figure A'B'C' using the rule (x, y) → (x−2, y+3).
Gnesinka [82]

Answer: B. (0, 8)

Step-by-step explanation:

The original point of B is (2, 5), being translated by the rule of (x-2, y+3). You will subtract 2 from the x value and add 3 to the y value. This will result in B’ (0, 8).

6 0
3 years ago
P(x)=Third-degree, with zeros of −3, −1, and 2, and passes through the point (1,12).
Mila [183]

Answer:

The polynomial is:

p(x) = -x^3 - 2x^2 + 5x + 6

Step-by-step explanation:

Zeros of a function:

Given a polynomial f(x), this polynomial has roots x_{1}, x_{2}, x_{n} such that it can be written as: a(x - x_{1})*(x - x_{2})*...*(x-x_n), in which a is the leading coefficient.

Zeros of −3, −1, and 2

This means that x_1 = -3, x_2 = -1, x_3 = 2. Thus

p(x) = a(x - x_{1})*(x - x_{2})*(x-x_3)

p(x) = a(x - (-3))*(x - (-1))*(x-2)

p(x) = a(x+3)(x+1)(x-2)

p(x) = a(x^2+4x+3)(x-2)

p(x) = a(x^3+2x^2-5x-6)

Passes through the point (1,12).

This means that when x = 1, p(x) = 12. We use this to find a.

12 = a(1 + 2 - 5 - 6)

-12a = 12

a = -\frac{12}{12}

a = -1

Thus

p(x) = -(x^3+2x^2-5x-6)

p(x) = -x^3 - 2x^2 + 5x + 6

6 0
3 years ago
PLEASE HELP!!
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3 years ago
Are three points collinear
Scorpion4ik [409]
<span>Three or more points.. </span> Two points are trivially collinear since two points<span>determine a line.</span>
6 0
4 years ago
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