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Hatshy [7]
3 years ago
15

Match the equation with the step needed to solve it.

Mathematics
1 answer:
viva [34]3 years ago
8 0
1) Subtract 2m
2) Subtract m
3) Add 1
4) Subtract 2
5) Add 2
6) Subtract 1
You might be interested in
Evaluate. -1/4(d+1) &lt; 2<br> Please show how
sergejj [24]

Answer:

-1/4 (d+1)< 2

(d+1) × -1 < 2×4

(d+1) × -1 < 8

-d - 1 < 8

-d < 8 + 1

-d < 9

d > -9

d = { -8, -7, -6....., 0, 1, 2, 3, 4,....}

Step-by-step explanation:

brainliest plz

4 0
2 years ago
A farmer creates a rectangular pen using part of the wall of a barn for one side of the pen and a total of 130 feet of fencing f
Nat2105 [25]

The equation that describes the which gives the area of the pen, A, as a function of x, the length of fence parallel to the barn is A = 130x - x²

<h3>How to find area of a rectangle?</h3>

The pen is rectangular. Therefore,

area of a rectangle = lw

where

  • l = length
  • w = width

Therefore,

perimeter = 2w + l

perimeter = 2w + x

130 = 2w + x

130 - x = 2w

w = 65 - 1/ 2 x

A = xw

A = x(65 - 1/ 2 x)

A = 65x - 1 / 2 x²

Therefore, the function that represents A, the area of the pen is as follows:

A = 130x - x²

learn more on rectangle here: brainly.com/question/12685460

#SPJ1

8 0
2 years ago
I only need letter D
strojnjashka [21]

Answer:

its how much money he will get

Step-by-step explanation:

6 0
3 years ago
On a linear X temperature scale, water freezes at −115.0°X and boils at 325.0°X. On a linear Y temperature scale, water freezes
belka [17]

Answer:

The current temperature on the X scale is 1150 °X.

Step-by-step explanation:

Let is determine first the ratio of change in X linear temperature scale to change in Y linear temperature scale:

r = \frac{\Delta T_{X}}{\Delta T_{Y}}

r = \frac{325\,^{\circ}X-(-115\,^{\circ}X)}{-25\,^{\circ}Y - (-65.00\,^{\circ}Y)}

r = 11\,\frac{^{\circ}X}{^{\circ}Y}

The difference between current temperature in Y linear scale with respect to freezing point is:

\Delta T_{Y} = 50\,^{\circ}Y - (-65\,^{\circ}Y)

\Delta T_{Y} = 115\,^{\circ}Y

The change in X linear scale is:

\Delta T_{X} = r\cdot \Delta T_{Y}

\Delta T_{X} = \left(11\,\frac{^{\circ}X}{^{\circ}Y} \right)\cdot (115\,^{\circ}Y)

\Delta T_{X} = 1265\,^{\circ}X

Lastly, the current temperature on the X scale is:

T_{X} = -115\,^{\circ}X + 1265\,^{\circ}X

T_{X} = 1150\,^{\circ}X

The current temperature on the X scale is 1150 °X.

5 0
3 years ago
Simplify the expression -8-4(-9-8v+3v)
Elanso [62]
Your answer is 108+60v.

5 0
3 years ago
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