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MA_775_DIABLO [31]

3 years ago

7

Factor each polynomial 6m^2-15m

1 answer:

stellarik [79]3 years ago

6 0

You can do prime factorization then find common factors to factor out.

6m² - 15m

2·3·m·m - 3·5·m

Both terms have 3·m so we can factor it out:

3·m (2·m - 5)

So that would be 3m (2m - 5)

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Can 0.2 0.6 3 and 9 make 24?

saveliy_v [14]

Answer:

Yes

Step-by-step explanation:

9 x 3 - (0.6/0.2) Simplify the parentheses

9 x 3 - 3 Multiply 9 by 3

27 - 3 Subtract

24

5 0

3 years ago

Setler79 [48]

To find equivalent equations, you simply have to simplify both of these :)

a) 5/6(x - 6) + 1/6(3 - x)

Simplify.

5/6x - 5 + 1/2 - 1/6x

Add like terms.

(5/6x - 1/6x) + (-5 + 1/2)

Simplify.

4/6x +(-4.5)

Simplify.

2/3x - 4.5

b) (y + 7)4 + 8 - 5y

Simplify.

4y + 28 + 8 - 5y

Add like terms.

(4y - 5y) + (28 + 8)

Simplify.

-y + 36

~Hope I helped!~

4 0

3 years ago

What is the expanded form of the decimal 7664.23? A) (7 x 100) + (6 x 10) + (6 x 1) + (4 x 1 10 ) Eliminate B) (7 x 100) + (6 x

Firlakuza [10]

Answer:

D

Step-by-step explanation:

D) (7 x 1000) + (6 x 100) + (6 x 10) + (4 x 1) + (2 x 1 10 ) + (3 x 1 100 )

3 0

3 years ago

Which equation will create a system of equations with infinitely many solutions?

SashulF [63]

Answer:

A and D

Step-by-step explanation:

for infinitely many solutions, ratio of coefficients should be equal.

since ratio = 1/4 so they will have infinitely many solutions.

6 0

3 years ago

PLZ HELP ASAP! ILL MARK BRAINLEST!

Sav [38]

Question 1.) Factor completely $12x^{4}+ 6x^{3}+18x^{2}$

Since $6$ divides all the three terms of the equation, so we take $6$ common out from the equation.

i.e., $6(2x^{4}+ 1x^{3}+3x^{2})$

Now, since $x^{2}$ divides all the three terms of the equation, so we take $x^{2}$ common out from the equation.

i.e., $6x^{2}(2x^{2}+ 1x+3)$

Now, this cannot be factored further as $2x^{2}+ 1x+3$ is a quadratic equation and taking out discriminant of it is:

$D = (1)^2 - 4(2)(3)\\ \ \ = 1- 24\\ \ \ = -23$

Since, the discriminant is less than zero. Therefore, the factored form of the given equation : $12x^{4}+ 6x^{3}+18x^{2}$ is : $6x^{2}(2x^{2}+ 1x+3)$

Question 2.) $7x^{3}y+ 14x^{2} y^{3} - 7x^{2}y^{2}$

Since $7$ divides all the three terms of the equation, so we take $7$ common out from the equation.

i.e., $7(x^{3}y+ 2x^{2} y^{3} - x^{2}y^{2})$

Now, since $x^{2}y$ divides all the three terms of the equation, so we take $x^{2}y$ common out from the equation.

i.e., $7x^{2}y(x+ 2y^{2} - y)$

Therefore, the factored form of the given equation : $7x^{3}y+ 14x^{2} y^{3} - 7x^{2}y^{2}$ is : $7x^{2}y(x+ 2y^{2} - y)$

Question 3.) What is the Greatest Common Factor of $x^{6}$ and $x^{3}$ ?

The Greatest Common Factor of two numbers is the biggest number which divides both the given numbers completely.

Now, for the expressions $x^{6}$ and $x^{3}$ :

Since, $x^{3}$ divides both the expressions completely and it is the biggest expression which divides both the expressions.

Therefore, $x^{3}$ is the Greatest Common Factor of $x^{6}$ and $x^{3}$

7 0

3 years ago

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