All categories

2 years ago

Factor completely 4а^2 - 8а - 5

Mathematics

1 answer:

Evgesh-ka [11]2 years ago

5 0

Answer:

(2a + 1)(2a - 5)

Step-by-step explanation:

Given

4a² - 8a - 5

Consider the factors of the product of the coefficient of the a² term and the constant term which sum to give the coefficient of the a- term

product = 4 × - 5 = - 20 and sum = - 8

The factors are + 2 and - 10

Use these factors to split the a- term

4a² + 2a - 10a - 5 ( factor the first/second and third/fourth terms )

= 2a(2a + 1) - 5(2a + 1) ← factor out (2a + 1) from each term

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

You might be interested in

Solve the system of equations. y+4z=0 4x+y-2z=2 -3x-3z=-9

antoniya [11.8K]

Step-by-step explanation:

y+4z=0

4x+y-2z=2

-3x-3z=-9

3 0

1 year ago

Read 2 more answers

400 A whole number a integer a real number or rational number

bearhunter [10]

Answer:

All of the above

Step-by-step explanation:

A whole number and integers are numbers that don't have decimals.

A rational number is a number that can be written as a fraction

A real number is any number that is not complex, which means that it has an imaginary solution or imaginary number part of it.

7 0

3 years ago

Legit question what is 10x 10 100 or 1000 I keep getting confused isn't it 100?

s344n2d4d5 [400]

Yea it is 100 :)

10x10=100

7 0

3 years ago

Read 2 more answers

I want the answers because i don't know how to do it

Andrew [12]

Question one is y( Insert sign) -9

6 0

3 years ago

Read 2 more answers

The angle θ 1 is located in Quadrant IV, and cos ⁡ ( θ 1 ) = 9/ 19 , theta, start subscript, 1, end subscript, right parenthesis

Firdavs [7]

Answer:

$sin\theta_1 = - \frac{2\sqrt{70}}{19}$

Step-by-step explanation:

We are given that $\theta_1$ is in fourth quadrant.

$cos\theta_1$ is always positive in 4th quadrant and

$sin\theta_1$ is always negative in 4th quadrant.

Also, we know the following identity about $sin\theta$ and $cos\theta$ :

$sin^2\theta + cos^2\theta = 1$

Using \theta_1 in place of \theta:

$sin^2\theta_1 + cos^2\theta_1 = 1$

We are given that $cos\theta_1 = \frac{9}{19}$

$\Rightarrow sin^2\theta_1 + \dfrac{9^2}{19^2} = 1\\\Rightarrow sin^2\theta_1 = 1 - \dfrac{81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{361-81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{280}{361}\\\Rightarrow sin\theta_1 = \sqrt{\dfrac{280}{361}}\\\Rightarrow sin\theta_1 = +\dfrac{2\sqrt{70}}{19}, -\dfrac{2\sqrt{70}}{19}$

$\theta_1$ is in 4th quadrant so $sin\theta_1$ is negative.

So, value of $sin\theta_1 = - \frac{2\sqrt{70}}{19}$

6 0

3 years ago