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Mrrafil [7]
2 years ago
8

I need help with this question please

Mathematics
1 answer:
11Alexandr11 [23.1K]2 years ago
3 0
Since you know that the equation of straight line is y=Mx + c
Where m mean gradient , and c means intercept
We don’t know our intercept neither do we know our gradient
So we have to find these to get our equation
Therefore, to find m( gradient) from the table
We are going to use the formulae for gradient which is
M=y2-y1 / x2-x1
I.e from the table
M= 23-15 / 4-2
M=8/2
M=4
Therefore how gradient is 4
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When y=_7, find the value of 2y2- 9
topjm [15]

2x(-7)^2-9 = 2x49-9=98-9=89 maybe

7 0
3 years ago
Simplify the following:<br> 3÷4i 15÷2-3i
seropon [69]

Answer:

7i=4.5

Step-by-step explanation:

Hope this helps

7 0
3 years ago
You accept a new job with a starting salary of $47,000. You receive a 4% raise at the start of your second year, a 5.6% raise at
Sunny_sXe [5.5K]

Answer:

1yr  47,000, 2nd yr  48,880, 3rd yr   51,617.28, 4th yr  57,346.7981 or 57,346.80

Step-by-step explanation:

1yr        47,000

2nd yr  47,000 x 4% = 48,880

3rd yr   48,880 x 5.6% = 51,617.28

4th yr   51,617.28 x 11.1% = 57,346.7981 or 57,346.80

4 0
2 years ago
The numbers -5,7,-18,and 0 are examples of
olga_2 [115]

Answer:

real numbers

Step-by-step explanation:

3 0
3 years ago
Note: Enter your answer and show all the steps
Brilliant_brown [7]

Answer: 1763

Step-by-step explanation:

Split this shape into three spaces, to where there are two square-like shapes above one rectangular shape. Next you will find the area of all three separate shapes and add them up to get your total area.

23m times 12m = 276,

23m times 12m = 276,

31m - 12m = 19m, (Here, we are taking the amounts we know of our rectangular shape to get the formula L times W used for determining this shape's area. We use the length of one of the squares(12m) subtracted by the length of both the rectangular shape and one of the squares.(31m) What we get is the lone value of L.

69m times 19m = 1311.

Add 276 + 276 + 1211. Answer = 1763.

5 0
3 years ago
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