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ruslelena [56]
2 years ago
8

Please answer this correctly

Mathematics
2 answers:
ivanzaharov [21]2 years ago
7 0
1-2^3-3^2
1-8-9
-7-9
-16
inessss [21]2 years ago
4 0

Answer:

-8+9 = 1

Step-by-step explanation:

-2 to the power of 3 = -8

3 to the power of 2 = 9

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Please help, I need help.<br><br>I don't understand.
Lana71 [14]

Find slope of both

#1

  • (11,2)
  • (22,4)

\\ \tt\hookrightarrow m=\dfrac{4-2}{22-11}=\dfrac{2}{11}=1.8

#2

  • (0,0)
  • (5,1)

\\ \tt\hookrightarrow m=\dfrac{1-0}{5-0}=\dfrac{1}{5}=0.2

Option C is correct

7 0
2 years ago
How are all angles in the 4th quadrant, negative or positive?
anastassius [24]

Answer:

They are all positive because you must go right.

They are all negative because you must go down.

Step-by-step explanation:

Example. (1,-4)

You go over one.

Down 4

5 0
3 years ago
Which expression is equivalent to (7x – 5) – (3x - 2)?
Alex73 [517]

Answer:D

Step-by-step explanation:

4 0
3 years ago
HELP PLEASE I WILL MARK YOU BRAINLIEST Jermias buys 6.5 pounds of ground beef at the grocery store. Ground beef costs $5.50 for
Svetlanka [38]

Answer: $35.75

Step-by-step explanation: 6.5x5.5=35.75

4 0
3 years ago
Read 2 more answers
Determine which function has the greatest rate of change over the interval [0, 2].
ruslelena [56]
Remember that the average rate of change of a function over an interval is the slope of the straight line connecting the end points of the interval. To find those slopes, we are going to use the slope formula: m= \frac{y_{2}-y_{1}}{x_2-x_1}

Rate of change of a:
From the graph we can infer that the end points are (0,1) and (2,4). So lets use our slope formula to find the rate of change of a:
m= \frac{y_{2}-y_{1}}{x_2-x_1}
m= \frac{4-1}{2-0}
m= \frac{3}{2}
m=1.5
The average rate of change of the function a over the interval [0,2] is 1.5

Rate of change of b:
Here the end points are (0,0) and (2,2)
m= \frac{2-0}{2-0}
m= \frac{2}{2}
m=1
The average rate of change of the function b over the interval [0,2] is 1

Rate of change of c:
Here the end points are (0,-1) and (2,0)
m= \frac{0-(-1)}{2-0}
m= \frac{1}{2}
m=0.5
The average rate of change of the function c over the interval [0,2] is 0.5

Rate of change of d:
Here the end points are (0,0.5) and (2,2.5)
m= \frac{2.5-0.5}{2-0}
m= \frac{2}{2}
m=1
The average rate of change of the function d over the interval [0,2] is 1

We can conclude that the <span>function that has the greatest rate of change over the interval [0, 2] is the function a.</span>
4 0
3 years ago
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