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Alex_Xolod [135]
4 years ago
6

Plz Simplify 5a-(3a-7)

Mathematics
1 answer:
lapo4ka [179]4 years ago
5 0
5a-(3a-7)
Equals
7+ 2a
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Solve each inequality and graph its solution. Can you guys solve all three please. This is do soon so please help!
mart [117]
<h2>Hi ☘</h2>

1-A 2-A 3-B

- 104 \geqslant  - 8b - 8

put -8 near -104

- 104 + 8 \geqslant  - 8b

- 96 \geqslant  - 8b

b will be alone.

\frac{ - 96}{ - 8}   \geqslant   \frac{ - 8b}{ - 8}

b  \geqslant 12

Answer A

3m - 7 >  - 4

3m >  - 4 + 7

3m > 3

\frac{3m}{3}  >  \frac{3}{3}

m > 1

Answer A

İf

m \geqslant 1

The answer would be D

But answer A

- 4x + 4 >  - 16

16 + 4  >  + 4x

20 > 4x

\frac{20}{4}  >  \frac{4x}{4}

x < 5

Answer B

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3 0
3 years ago
Jk is perpendicular to xy at its midpoint m. choose the answer that includes the correct responses.
Soloha48 [4]

Answer:

D

Step-by-step explanation:

JK intercept line XY at the midpoint. This means that these points are all equidistant from the midpoint. Due to this, J and K are both an equal distance from M. So is X and Y.

3 0
3 years ago
The table above gives values of the differentiable functions f and g, and f', the derivative of f, at selected values of x. If g
Aleks [24]

Answer:

B

Step-by-step explanation:

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7 0
2 years ago
On a coordinate plane, parallelogram H I J K is shown. Point H is at (negative 2, 2), point I is at (4, 3), point J is at (4, ne
Mrrafil [7]
<h3>Answer:  Choice C.   (1,0)</h3>

============================================================

Explanation:

Point H is at (-2,2) and J is at (4,-2)

Focusing on the x coordinates, the midpoint is (x1+x2)/2 = (-2+4)/2 = 1.

The y coordinates have a midpoint of (y1+y2)/2 = (2+(-2))/2 = 0.

The midpoint is at (1,0). That's why the answer is choice C.

--------------------------

This next section is optional.

Points I and K are located at (4,3) and (-2,-3) respectively

Averaging the x coordinates gets us (x1+x2)/2 = (4+(-2))/2 = 1

Doing the same for the y coordinates gets us (y1+y2)/2 = (3+(-3))/2 = 0

We end up with the same midpoint (1,0)

The midpoint being the same for both diagonals proves that each diagonal is bisected, ie cut in half.

3 0
3 years ago
WILL GIVE BRAINLIEST (NO LINKS ALLOWED)
Setler [38]
B pretty sure that it’s that
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