Ask question

Ask question

All categories

2 years ago

5

The greatest common factor of 32a and 48b

1 answer:

rodikova [14]2 years ago

7 0

The common factors of 32 and 48 are 16, 8, 4, 2, 1,

You might be interested in

How to find the θ in this equation: tan(θ+16°)=1/tan(θ+60°) by using the trigonometric identities?thanks

DedPeter [7]

Answer:

$\huge \theta =7°$

Step-by-step explanation:

$tan( \theta + 16 \degree) = \frac{1}{tan( \theta + 60 \degree) } \\ tan( \theta + 16 \degree) =cot( \theta + 60 \degree) \\ tan( \theta + 16 \degree) = tan \{90 \degree - (\theta + 60\degree) \} \\ \theta + 16 \degree = 90 \degree - \theta - 60\degree \\ \theta +\theta = 30 \degree - 16\degree \\2 \theta = 14 \degree \\ \theta = \frac{14 \degree}{2} \\ \huge \red{ \boxed{\theta =7 \degree}}$

7 0

3 years ago

What is 1.147 , 1.15 , 1.098 , 0.884 from greatest to least

SOVA2 [1]

1.15,1.147,1.098,.884

5 0

3 years ago

Read 2 more answers

An expression is shown below:

Alex73 [517]

Given:

The given expression is:

$6x^2y-3xy-24xy^2+12y^2$

To find:

Part A: The expression by factoring out the greatest common factor.

Part B: Factor the entire expression completely.

Solution:

Part A:

We have,

$6x^2y-3xy-24xy^2+12y^2$

Taking out the highest common factor 3y, we get

$=3y(2x^2-x-8xy+4y)$

Therefore, the required expression is $3y(2x^2-x-8xy+4y)$ .

Part B:

From part A, we have,

$3y(2x^2-x-8xy+4y)$

By grouping method, we get

$=3y(x(2x-1)-4y(2x-1))$

$=3y(x-4y)(2x-1)$

Therefore, the required factored form of the given expression is $3y(x-4y)(2x-1)$ .

3 0

2 years ago

Simplify: a + 4b+9 +6a +2a

Anit [1.1K]

Answer:

9a + 9 + 4b

Have a wonderful day! :D

3 0

3 years ago

Read 2 more answers

Somebody link me a video on how to do proofs

Artyom0805 [142]

Https://youtu.be/fSu1LKnhM5Q

7 0

2 years ago