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AURORKA [14]
2 years ago
11

Danielle has her finals coming up and she's worried she'll fail her class. Her past 3 test grades were a 72. an 85, and a 68. Sh

e needs a 70 to pass the class. What grade does she need to make on the final to get a 70?
Mathematics
1 answer:
natka813 [3]2 years ago
7 0

Answer:

55

Step-by-step explanation:

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FACTOR EACH EXPRESSION, IF IT CANT BE FACTORED WRITE CANNOT BE FACTORED. 1. 4x+12
dimaraw [331]
1.) 4(x+3)
Find the GCF, Greatest Common Factor, of 4x and 12. 
4x=2*2*x
12=3*2*2
The greatest common factor is 4. Put this outside of the parentheses.             (You would multiply the 2*2)
Then, put the rest of the factors as a sum. (Only the factors on the same line.)
Solution: 4(x+3)

To check, distribute to see if it works.

4x+12

2.) 2(4r+7)
Find the GCF of 8r and 14
8r=2*2*2*r
14= -1*7*2
The greatest common factor is 2.                                                                 (There is only 1 two, so you would not multiply them.)
Then, put the rest of the factors as a sum. (Only the factors on the same line.)
Multiply the 2*2*r as one addend and the -1*7 as the other.
Solution: 2(4r-7)

To check, distribute to see if it works.

8r-14

Do you get it now?

3.) 5(x+7)
4.) 7(2x+1)
5.) Cannot be factored.

32x-15
Find the GCF of 32x and -15

32x: 2*2*2*2*2*x
-15: -1*5*3

Because there are no similar factors other than 1, it cannot be factored.

6.)  8(4x+3)
7.) 3(2x-3)
8.) 24(1x+2)
9.) 9(-2x+8)
10.) Cannot be factored
11.) 8(1x+3)
12.) 50(1x+5)



4 0
3 years ago
Read 2 more answers
Someone please help quickly :(
postnew [5]
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8 0
2 years ago
Read 2 more answers
Determine whether the integral is convergent or divergent. ∫[infinity] 2 e^−1/x / x^2 dx : O Convergent O divergent If it is con
monitta

Let f(x)=e^{-1/x}. Then f'(x)=\frac1{x^2}e^{-1/x}>0 for all x\ge2, so f is strictly increasing. As x\to\infty, e^{-1/x}\to e^0=1, so f is bounded above by 1. This is to say,

e^{-1/x}

and the integral of \frac1{x^2} converges over the same domain, so this integral must also converge by comparison.

We have, by setting y=-\frac1x,

\displaystyle\int_2^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx=\int_{-1/2}^0e^y\,\mathrm dy=e^0-e^{-1/2}=1-\frac1{\sqrt e}

8 0
2 years ago
About 18% of the population of a large country is nervous around strangers
sashaice [31]

Answer:

ok

Step-by-step explanation:

4 0
2 years ago
PLEASE ANSWER ASAP
professor190 [17]

Answer:

hmmm I think u need to talk to ur teacher

7 0
2 years ago
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