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3 years ago

Find the Greatest Common Factor of 3a^2+9

Mathematics

2 answers:

AlladinOne [14]3 years ago

7 0

Answer:

Step-by-step explanation:

We are given that an expression

$3a^2+9$

We have to find the greatest common factor of $3a^2+9$

In order to find the GCF we will find the factor of each term

$3a^2=3\times a\times a$

$9=3\times 3$

The common factor in both terms is 3.

Hence, greatest common factor of $3a^2+9$ is 3.

Answer: 3

Nataly [62]3 years ago

6 0

The GCF of 3a^2 + 9 is 3.

Factor for 3a^2 + 9
3a^2 + 9 = 3(a2 + 3)

Answer:3(a2+3)

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15% off the original price of $350, plus 6% sales tax

jonny [76]

Step 1. 100-15=85
Step 2. $350 *.85=$297.50
Step 3. $297.50 * .06= $17.85
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Answer: $315.35

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3 years ago

What factors multiply to -80 and add to 6

forsale [732]

80 = -1(80), -2(40), -4(20), -5(4),

Total answers include: -81, 79, -42, 16, -9, and 1.

None of them add up to 6!

Are you doing polynomial equations?

If so, you can solve it the hard way.

x^2 + 6x -80 = 0

(x^2 + 6x + 9) - 80 = 9

(x+3)^2 = 89

(x+3) = 9.43

x = 6.43

If you're not doing those then I'm afraid your question has no answer.

Hope that helps!

7 0

3 years ago

A bag contains different colored candies. There are 50 candies in the bag, 28 are red, 10 are blue, 8 are green and 4 are yellow

Mrrafil [7]

Answer:

$\displaystyle \frac{54}{5405}$ .

Step-by-step explanation:

How many unique combinations are possible in total?

This question takes 5 objects randomly out of a bag of 50 objects. The order in which these objects come out doesn't matter. Therefore, the number of unique choices possible will the sames as the combination

$\displaystyle \left(50\atop 5\right) = 2,118,760$ .

How many out of that 2,118,760 combinations will satisfy the request?

Number of ways to choose 2 red candies out a batch of 28:

$\displaystyle \left( 28\atop 2\right) = 378$ .

Number of ways to choose 3 green candies out of a batch of 8:

$\displaystyle \left(8\atop 3\right)=56$ .

However, choosing two red candies out of a batch of 28 red candies does not influence the number of ways of choosing three green candies out of a batch of 8 green candies. The number of ways of choosing 2 red candies and 3 green candies will be the product of the two numbers of ways of choosing

$\displaystyle \left( 28\atop 2\right) \cdot \left(8\atop 3\right) = 378\times 56 = 21,168$ .

The probability that the 5 candies chosen out of the 50 contain 2 red and 3 green will be:

$\displaystyle \frac{21,168}{2,118,760} = \frac{54}{5405}$ .

3 0

3 years ago

The bill for the repair of a computer was $179. The cost of part was $44, and labor charge was $45 per hour. How many hours did

Cloud [144]

Alright! First you'd subtract the amount of the part ($44) from your total ($179), then you'd divide by the rate of cost ($45). Your equation would be:
($179 - $44) ÷ $45 = h.

Then you solve:
($179 - $44) = $135
($135 ÷ $45) = 3
h = 3.

The answer is 3 hours.

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Travka [436]

You can do this: 71,000 or 71,000/1

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