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2 years ago

Factor completely 3x^2+9x+6

Mathematics

1 answer:

anastassius [24]2 years ago

7 0

Answer:

Step-by-step explanation:

3x²+9x+6

=3(x²+3x+2)

=3[x²+x+2x+2]

=3[x(x+1)+2(x+1)]

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

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a park is 4 times as long as it is wide. if the distance around the park is 12.5 kilometers, what is the area of the park?

blsea [12.9K]

So, first, what are the two sides?

let's call then x and y

we know that 2(x+y)=12.5 (that's the distance around)

so that means that x+y=6.25 (I just divided both by 2)

now, x=4y (from "4 times as long as it is wide")

so we can substitute:

x+4x=6.25

5x=6.25

x=1.25

so one side, is 1.25 and the other will be 1.25*4=5

and for the area we multiply the two:

1.25*5=6.25 square kilometers, and this is the answer!

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2 years ago

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How do you compare 7/8 ___ 15/16 ?

solong [7]

Answer:

7/8 < 15/16

Step-by-step explanation:

Get a common denominator of 16

7/8 *2/2

14/16

14/16 < 15/16

so 7/8 < 15/16

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2 years ago

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Step-by-step explanation:

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2 years ago

How many 5-member chess teams can be chosen from 15 interested players? Consider only the members selected, not their board posi

ryzh [129]

<u>Answer</u>:

3003 number of 5-member chess teams can be chosen from 15 interested players.

<u>Step-by-step explanation:</u>

Given:

Number of the interested players = 15

To Find:

Number of 5-member chess teams that can be chosen = ?

Solution:

Combinations are a way to calculate the total outcomes of an event where order of the outcomes does not matter. To calculate combinations, we will use the formula $nCr = \frac{n!}{r!(n - r)!}$

where

n represents the total number of items,

r represents the number of items being chosen at a time.

Now we have n = 15 and r = 5

Substituting the values,

$15C_5 = \frac{15!}{5!(15- 5)!}$

$15C_5 = \frac{15!}{5!(10)!}$

$15C_5 = \frac{15\times \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 }{5 \times 4 \times 3 \times 2 \times 1(10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}$

$15C_5 = \frac{15\times \times 14 \times 13 \times 12 \times 11}{(5 \times 4 \times 3 \times 2 \times 1)}$

$15C_5 = \frac{360360}{120}$

$15C_5 = 3003$

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2 years ago