All categories

2 years ago

Factor 2x2 -5x - 3 Group of answer choices (2x - 1) ( x + 3 ) (x - 1) ( x + 3 ) (2x +1) ( x - 3 ) (x + 1) ( x - 3 )

Mathematics

1 answer:

Marrrta [24]2 years ago

4 0

Answer:

2x2-5x-3

we will take 6's factors that would be the together the sum 5

2x2-6x+1x-3

2x(x-3) +1(x-3)

(2x+1) (x-3)

You might be interested in

Answer 5, 6 , and 7 by today please.

kherson [118]

Answer:

Step-by-step explanation:

5. a) ∠1 and ∠2 are remote interior angles of ∠ACD so that means that ∠ACD = ∠1 + ∠2

b) Because an exterior angle is the sum of its two remote interior angles it makes sense that an exterior angle is greater in measure than either of its remote interior angles.

6. BD = DB Reflexive property

∠3 = ∠5, ∠4 = ∠6 Alt. int. angles

ΔADB = ΔCDB ASA

7. AB = BC Def. of midpoint

∠1 = ∠2 Given

∠BAE = ∠CBD Corresponding angles

ΔBAE = ΔCBD ASA

∠D = ∠E CPCTC

3 0

2 years ago

If 1/8 pound of fertilizer covers 25square feet of ground how much fertilizer in pounds is needed to cover 1 square foot of grou

slamgirl [31]

Answer:

0.005

Step-by-step explanation:

(1/8) /25

8 0

2 years ago

Read 2 more answers

Is this a direct variation problem

Llana [10]

Yes and for a. You can just put y=x [y=1x is bascually the same but just the 1 isnt needed]

6 0

3 years ago

The expression 3(x-9) is equivalent to<br> 3(x)+9.<br> 3(x)+3(9).<br> 3(x) – 9.<br> 3(x) – 3(9).

sukhopar [10]

Answer:

3(x) – 3(9)

Step-by-step explanation:

3(x-9)

Distribute the 3 to each term in the parentheses

3*x -3*9

3x -27

3 0

2 years ago

Read 2 more answers

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago