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2 years ago

I NEED HELP NOW PLS………………………………………..pdkxjxj

Mathematics

1 answer:

jolli1 [7]2 years ago

8 0

Answer:

The correct answer is D the last option.

Step-by-step explanation:

hope it helps brainliest pls

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If f(x) = x2 - 2x and g(x) = 6x + 4 for which value of x does (t +g)(x)=0

Vinil7 [7]

Answer:

-2

Assumption:

Find the value of x such that $(f+g)(x)=0$ .

Step-by-step explanation:

$(f+g)(x)=0$

$f(x)+g(x)=0$

$(x^2-2x)+(6x+4)=0$

Combine like terms:

$x^2+4x+4=0$

This is not too bad too factor on the left hand side since 2(2)=4 and 2+2=4.

$(x+2)(x+2)=0$

$(x+2)^2=0$

So we need to solve:

$x+2=0$

Subtract 2 on both sides:

$x=-2$

Let's check:

$(f+g)(-2)$

$f(-2)+g(-2)$

$((-2)^2-2(-2))+(6(-2)+4)$

$(4+4)+(-12+4)$

$(8)+(-8)$

$0$

0 was the desired output of $(f+g)(x)$ .

7 0

2 years ago

What is the factorization of the polynomial below?<br> 9x2 - 16

Agata [3.3K]

use (a-b) (a+b )

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

8 0

2 years ago

Read 2 more answers

What is (13x + 9k) + (17x + 6k), simplified?

Mazyrski [523]

The answer is 15x+30k

7 0

2 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 =

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