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Rainbow [258]
3 years ago
14

Factor this expression. Select the correct answer from each drop-down menu.

Mathematics
2 answers:
umka21 [38]3 years ago
7 0
The answer is (3x+4)(x-5)
ExtremeBDS [4]3 years ago
7 0

Answer:

2 and 5

Step-by-step explanation:

Use the sum-product pattern

3x^{2}{\color{#c92786}{-11x}}-20

3x^{2}+{\color{#c92786}{4x}}{\color{#c92786}{-15x}}-20

2

Common factor from the two pairs

3x^{2}+4x-15x-20

x(3x+4)-5(3x+4)

3

Rewrite in factored form

x(3x+4)-5(3x+4)

(x-5)(3x+4)

Solution

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Kaylis [27]

Answer:

option A is the answer hope you find it helpful

3 0
3 years ago
How to solve this :') please help
blsea [12.9K]

Answer:

3/2

Step-by-step explanation:

sin(3π/4 - β) = sin(3π/4)cosβ - cos(3π/4)sinπ =

\sin(\frac{3\pi}{4}-\beta)=\sin(\frac{3\pi}{4})\cos\beta-\cos\frac{3\pi}{4}\sin\beta\\=\frac{1}{\sqrt{2}}\cos\beta-(-\frac{1}{\sqrt{2}})\sin\beta\\\\=\frac{1}{\sqrt{2}}(\cos\beta +\sin\beta)\\\\\sin^2(\frac{3\pi}{4}-\beta)=\frac{1}{2}\cos\beta +\sin\beta)^2=\frac{1}{2}(\cos^2\beta +\sin^2\beta+2\sin\beta\cos\beta\\=\frac{1}{2}(1+\sin2\beta)=\frac{1}{2}(1-\frac{1}{5}) = \frac{2}{5}\\

Use \cot^2x=\csc^2x-1=\frac{1}{\sin^2x}-1

so

\cot^2(\frac{3\pi}{4}-\beta)=\frac{1}{\sin^2(\frac{3\pi}{4}-\beta)}-1 = \frac{1}{\frac{2}{5}}-1=\frac{5}{2}-1=\frac{3}{2}

7 0
2 years ago
Luisa solves for x in the equation 9(1/3) + 8x = 4( 2x + 3/4)
Maru [420]

Answer:

x = all real numbers because this equation is an identity

Step-by-step explanation:

9 (1/3) = 9*1/3 = 3

3 + 8x = 4(2x + 3/4)

3 + 8x = 8x + 3

x = all real numbers because this equation is an identity

7 0
3 years ago
Read 2 more answers
Which of the following is a conjugate for 7 + i Square root of 2? (2 points) − 7 + i Square root of 2 7 − i Square root of 2 i S
hammer [34]

Answer:

7-\sqrt{2}i

Step-by-step explanation:

Given number is 7+\sqrt{2}i. We need to find the complex conjugate of the given number.

We will use the general rule to find out the complex conjugate of the number.

The complex conjugate of a+bi will be a-bi

And the complex conjugate of 7+\sqrt{2}i\ became\ 7-\sqrt{2}i

So, the complex conjugate of 7+\sqrt{2}i is 7-\sqrt{2}i

3 0
3 years ago
PLZ ANSWER THIS QUESTION <br><br> what is 10 decreased by u.
tigry1 [53]

Answer:

10-u

Step-by-step explanation:

7 0
3 years ago
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