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Margarita [4]
3 years ago
10

I really need help they never taught me this in my old school.

Mathematics
1 answer:
SIZIF [17.4K]3 years ago
6 0
To find this, just plug the numbers into the expressions.

5x = y
5(2) = 7
10 = 7
So it can't be A.

x + 2 = y
2 + 2 = 7
4 = 7

x + 4 = y
2 + 4 = 7
6 = 7

2x + 3 = y
2(2) + 3 = 7
7 = 7

The correct answer is
D. 2x + 3
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Which of the following g expressions is equal to 3x^2+27
Dafna1 [17]

Answer:

Answer is 3(x^2 + 9), when factored.

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
To win a game you need at least 45 points.Each question is worth 3 points.Write and solve an inequality that represents the numb
KonstantinChe [14]
You need 45
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3 points each question correct
The answer is 3x less than or equal to 45
3 0
4 years ago
1) Find an equation of the line that passes through thr pair of points (-5,-9) and (2,-7).
Yuki888 [10]
1: Assuming this line is linear, the first thing to do is find the slope. 
The slope formula is m=\frac{ y_{2} - y_{1} }{ x_{2} - x_{1} }, so in this case, it is m=\frac{ (-7) - (-9) }{ 2 - (-5) }, which comes to \frac{2}{7}. Then, use your slope and one of the points given to write the point-slope formula of the linear equation, and then written in standard form:
y=\frac{2}{7}x-\frac{53}{7}.

2. Use the same process as in question 1 to find:
y=-7x+11


8 0
4 years ago
Given circle N with chord AB, which point lies on the perpendicular bisector of AB?
aivan3 [116]

The point that lies on the perpendicular bisector of AB is point N

<h3>How to determine the point?</h3>

The circle is given as:

Circle N

The chord is given as:

Chord AB

The perpendicular bisector of the chord is any line drawn from the center of the circle that divides the chord in equal halves.

This means that the point starts from point N

Hence, the point that lies on the perpendicular bisector of AB is point N

Read more about perpendicular bisectors at:

brainly.com/question/929137

#SPJ1

3 0
2 years ago
-2/3(3 2/3) I'm not sureeeee
zalisa [80]
2 11/25





Hope I helped!
5 0
3 years ago
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