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3 years ago

Factor the expression. 100k^3 – 75k^2 + 120k – 90

Mathematics

1 answer:

iren [92.7K]3 years ago

6 0

For this case we must factor the following expression:

$100k ^ 3-75k ^ 2 + 120k-90$

We take common factor 5:

$5 (20k ^ 3-15k ^ 2 + 24k-18) =$

We have two groups within the parentheses:

Group 1: $20k ^ 3-15k ^ 2$

Group 2: $24k-18$

We factor each group:

Group 1: $6 (4k-3)$

Group 2: $5k ^ 2 (4k-3)$

Rewriting we have:

$5 (5k ^ 2 (4k-3) +6 (4k-3)) =\\5 ((5k ^ 2 + 6) (4k-3))$

Answer:

$5 (5k ^ 2 + 6) (4k-3)$

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Solve 5y^2+20=80, where y is the real number

soldi70 [24.7K]

Answer:

2√3

Step-by-step explanation:

5y2+20=80

Step 1: Subtract 20 from both sides.

5y^2+20−20=80−20

5y^2=60

Step 2: Divide both sides by 5.

5y^2 /5 = 60/5

y2=12

Step 3: Take square root.

y=√12

y=2√3

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3 years ago

It’s right angle trig WILL MARK BRAINLIEST!!!

erastova [34]

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3 years ago

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The system of equations has no solution: y=2/3x-5 4x - 6y = 30 true or false?

Serjik [45]

1. y = 2/3x - 5

2. 4x - 6y = 30

Divide 2. by 2

3. 2x - 3y = 15

Substitute 1. into 3.

4. 2x - 3(2/3x - 5) = 15

5. 2x - 2x + 15 = 15

6. 15 = 15

False. There are an infinite number of solutions.

3 0

3 years ago

The equation of a line is given below . 6x+2y =-18 find the x intercept and the y intercept

vova2212 [387]

Answer: x intercept is -3

y intercept is -9

Step-by-step explanation: 6(0) = 2y=-18, divide both sides by 2, get -9 as a y intercept

6x+2(0)=-18

divide both sides by 6, and get -3 as an x intercept

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3 years ago

What degree of rotation is represented on this matrix

Korvikt [17]

Answer:

Option B is correct

the degree of rotation is, $-90^{\circ}$

Step-by-step explanation:

A rotation matrix is a matrix that is used to perform a rotation in Euclidean space.

To find the degree of rotation using a standard rotation matrix i.e,

$R = \begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$

Given the matrix: $\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$

Now, equate the given matrix with standard matrix we have;

$\begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$ = $\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$

On comparing we get;

$\cos \theta = 0$ and $-\sin \theta =1$

As,we know:

$\cos \theta = \cos(-\theta)$

$\sin(-\theta) = -\sin \theta$

$\cos \theta = \cos(90^{\circ}) = \cos( -90^{\circ})$

we get;

$\theta = -90^{\circ}$

and

$\sin \theta =- \sin (90^{\circ}) = \sin ( -90^{\circ})$

we get;

$\theta = -90^{\circ}$

Therefore, the degree of rotation is, $-90^{\circ}$

7 0

3 years ago