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postnew [5]
3 years ago
6

Please help!!! Will mark!!

Mathematics
1 answer:
yawa3891 [41]3 years ago
8 0
X^2+5x+6=0

x^2+2x+3x+6=0

x(x+2)+3(x+2)=0

(x+3)(x+2)=0

So the roots, zeros, or points where the graph touch the x-axis are when x=-2 and -3.


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Answer:

sin⁴x - sin²x

sin²x(sin²x - 1)

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(1 - cos²x)(-cos²x)

-cos²x + cos⁴x

cos⁴x - cos²x

4 0
3 years ago
Factor 10x^3 8x^2 - 6x. (<br> a.2x(5x 4x - 3) (<br> b.2x(5x 4x^2- 3) (<br> c.2x(5x^2 4x - 3)
Semmy [17]
You have to look at the equation and find a common factor bettween all parts. SO, we see they all have an x, so that is the first thing you can factor out. They also have 
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4 0
3 years ago
Can someone please help me??
iren2701 [21]

Answer:

the first one

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using y=mx+b, you can find which has a "bigger" slope.

8 0
3 years ago
Read 2 more answers
What is the solution to the linear equation?
Airida [17]
2x/3 - 1/2 = 1/3 - 5x/6
The common denominator is 2 * 3 = 6  Multiply through by 6
6(2x/3 - 1/2 = 1/3 - 5x/6)
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3 0
3 years ago
Read 2 more answers
Fill in the missing portions of the function to rewrite g(x) = 3a^2 − 42a + 135 to reveal the zeros of the function. What are th
mariarad [96]

Answer:

g(x) = 3(x-9)(x-5)

Zeros: x = 9 and x = 5.

Step-by-step explanation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this question:

g(x) = 3x^{2} - 42a + 135

So

a = 3, b = -42, c = 135

\bigtriangleup = (-42)^{2} - 4*3*135 = 144

x_{1} = \frac{-(-42) + \sqrt{144}}{2*3} = 9

x_{2} = \frac{-(42) - \sqrt{144}}{2*3} = 5

So

g(x) = 3(x-9)(x-5)

Zeros: x = 9 and x = 5.

8 0
4 years ago
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