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mestny [16]
3 years ago
12

Hi, this isn't a math question or anything. But I have a question about high schools. So I have all A's for my first marking per

iod and one C for an enirchnment. But do you think I could get into a good high school like Central? Also, I'm not taking the PSSA or big tests. Also, I'm in 7th grade if your wondering.
Mathematics
1 answer:
beks73 [17]3 years ago
6 0

Answer:

Depends on your GPA

Step-by-step explanation:

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linda scored 35, 47, and 42 points on three tests. how many points should she get on the fourth test to get an average score of
aliya0001 [1]
(35 + 47 + 42 + x) / 4 = 50
(124 + x) / 4 = 50....multiply both sides by 4, cancelling the 4 on the left side.
124 + x = 50 * 4
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3 years ago
Please help me IDK how to do this!!!!!!
tatiyna
A parallelogram should have 2 sets of parallel lines. Let's find the slope of line PQ and RS to test.

PQ:
(4-2)/(1-(-3))
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RS:
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Because 1 does not equal 1/2 (the slopes are different) the lines are not parallel. Thus, the figure is not a parallelogram.
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3 years ago
Find the values of k so that each remainder is three. <br> 10. (x^2+ 5x + 7) = (x + k)
Goryan [66]

Answer:

k=1\text{ or } k=4

Step-by-step explanation:

We can use the Polynomial Remainder Theorem. It states that if we divide a polynomial P(x) by a <em>binomial</em> in the form (x - a), then our remainder will be P(a).

We are dividing:

(x^2+5x+7)\div(x+k)

So, a polynomial by a binomial factor.

Our factor is (x + k) or (x - (-k)). Using the form (x - a), our a = -k.

We want our remainder to be 3. So, P(a)=P(-k)=3.

Therefore:

(-k)^2+5(-k)+7=3

Simplify:

k^2-5k+7=3

Solve for <em>k</em>. Subtract 3 from both sides:

k^2-5k+4=0

Factor:

(k-1)(k-4)=0

Zero Product Property:

k-1=0\text{ or } k-4=0

Solve:

k=1\text{ or } k=4

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(x^2+5x+7)\div(x+1)\text{ or } (x^2+5x+7)\div(x+4)

Will yield 3 as the remainder.

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3 years ago
Which is not a property of countercurrent multiplication?
laiz [17]

Answer:

It is opposed by the vasa recta.

Step-by-step explanation:

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3 years ago
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3 years ago
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