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2 years ago

Which of the following represents the factorization of the trinomial below?

Mathematics

1 answer:

DerKrebs [107]2 years ago

7 0

Answer:

A.(3x+1)*(2x+3)

Step-by-step explanation:

6x2+11x+3

= 6x2 + 9x +2x +3

= 3x*(2x+3) + (2x+3)

= (3x+1)*(2x+3)

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Which choice shows y=3x+2 and y=3x2 correctly paired with their graphs?

Blababa [14]

Answer:

The second one

7 0

2 years ago

If an integer m is divisible by both 10 and 12, then it must also be divisible by which of the following?

Murrr4er [49]

Answer:

Step-by-step explanation:

an integer m is divisible by both 10 and 12

Lets write the prime factor for each number

the prime factor for 10 is 5,2

the prime factor for 12 is 3,2,2

Now take the product of the prime factors that are not in common and also in common $5 \cdot 3 \cdot 2 \fdot 2= 60$

60 is divisible by 20

so answer is 20

3 0

2 years ago

A box-and-whisker plot. The number line goes from 0 to 9. The whiskers range from 0 to 7, indicated by A and E. The box ranges f

Sladkaya [172]

minimum value: 1

lower quartile: 3

median of the data: 6

upper quartile: 8

maximum value: 14

Sorry for the late answer, i hope this will help you in the future and everyone else who reads this.

6 0

2 years ago

A school purchased T-shirts to sell for a fundraiser. The school spent $684 for the T-shirts. Each T-shirt will sell for $12.

Setler79 [48]

Answer:

If the question is How many T-Shirts does the school have, it will be 57 T-Shirts

Step-by-step explanation:

684 divided by 12 is 57

8 0

2 years ago

Simplify the expression

chubhunter [2.5K]

Answer:

sin(α) -cos(α)

Step-by-step explanation:

$\dfrac{2\sin^2{\alpha}-1}{\sin{\alpha}+\cos{\alpha}}=\dfrac{(2\sin^2{\alpha}-1)(\sin{\alpha}-\cos{\alpha})}{(\sin{\alpha}+\cos{\alpha})(\sin{\alpha}-\cos{\alpha})}\\\\=\dfrac{(2\sin^2{\alpha}-1)(\sin{\alpha}-\cos{\alpha})}{\sin^2{\alpha}-\cos^2{\alpha}}=\dfrac{(2\sin^2{\alpha}-1)(\sin{\alpha}-\cos{\alpha})}{\sin^2{\alpha}-(1-\sin^2{\alpha})}\\\\=\dfrac{(2\sin^2{\alpha}-1)(\sin{\alpha}-\cos{\alpha})}{2\sin^2{\alpha}-1}\\\\=\sin{\alpha}-\cos{\alpha}$

4 0

3 years ago