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mina [271]
3 years ago
10

How do you find prime factorization?

Mathematics
1 answer:
skad [1K]3 years ago
6 0
You can use a factor tree, and if you dont know how to use one, look it up on youtube, theres tons
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-\frac{4\left(x+3\right)}{5}=4x-12
Keith_Richards [23]

Answer:

huh

Step-by-step explanation:

6 0
3 years ago
What is the slope of the line that passes through the points (1, 3) and (5, –2)?
Sergeu [11.5K]

m=-(5/4)

From left to right, (1,3) is first and then comes (5,-2). Always remember when finding slopes without equations, the rule is RISE over RUN, to the numerator and denominator, respectively.

The y value of the second coordinates becomes negative which is unlike the y value in the first coordinates, which means our slope is downward, meaning it has a negative sign in front.

In every slope, there’s a numerator, being the rise, and a denominator, being the run.

To find the rise, we must look at the y values. Starting at 3 going to -2 has a space of 5 units, making that our numerator.

To find the run, the first x value is 1 and the second is 5, making a space of 4, which is out denominator.

With these two numbers and the negative sign, we get -(5/4) as our slope.

7 0
3 years ago
∞(PLEASE HURRY)
slamgirl [31]

Answer:

1/2

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
In a bag of sweets, 6 are orange and 9 are lime. Jodi chooses a sweet at random, eats it an
Digiron [165]

Answer:\frac{17}{35}

Step-by-step explanation:

Given

The bag contains 6 orange and 9 lime sweets

For both sweets of the same flavor, Jodi must either choose orange or lime

No. of ways it can be done is ^6C_1\times^5C_1+^9C_1\times^8C_1

The total no of ways of selecting two sweets out of 15 is ^{15}C_1\times^{14}C_1

The probability that both sweets are the same flavor

P=\dfrac{^6C_1\times^5C_1+^9C_1\times^8C_1}{^{15}C_1\times^{14}C_1}=\frac{102}{210}=\frac{51}{105}=\frac{17}{35}

3 0
3 years ago
Given the sequence 8, 12, 16, 20, 24, what is the sum of the 31st and 19th terms?
erica [24]

Answer:

First term (a) =8

Common difference (d)= t2-t1

=12-8

=4

Now, sum of first 31th term (tn31) =n/2{2a+(n-1)d}

= 31/2{2×8+(31-1)4}

=31/2{16+(30×4)

=31/2(16+120)

=31/2×126

=31×63

Step-by-step explanation:

Similarly use 19 as (n) for the 19th term

7 0
3 years ago
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