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s344n2d4d5 [400]
3 years ago
15

Find the inverse of y=-12x+7

Mathematics
2 answers:
gayaneshka [121]3 years ago
7 0
You want to solve for x here.

When finding the inverse of a function, you might get this form of the equation:
f(x) = -12x + 7
And you would have to change the f(x) into a (y). However, this equation already has y in place of f(x), so you can just get started!

To find the inverse you want to solve for x, and, thus, you must ISOLATE x from every other term and variable.

1) (subtract 7 from each side)
y - 7 = -12x

2) (divide -12 from each side)
(y/-12) + (7/12) = x

Now that x is now isolated, you have your answer. Your teacher, however, might want you to change the (y)s in your answer to (x)s.

So the inverse of y = -12x + 7 is
(y/-12) + (7/12)

OR (if your teacher wants you to change your (y)s to (x)s in your inverse solution (which they probably do)):

The inverse of y = -12x + 7 is
(x/-12) + (7/12)
Oksana_A [137]3 years ago
5 0
1.1 Solve y-12x+7 = 0

Tiger recognizes that we have here an equation of a straight line. Such an equation is usually written y=mx+b ("y=mx+c" in the UK).

"y=mx+b" is the formula of a straight line drawn on Cartesian coordinate system in which "y" is the vertical axis and "x" the horizontal axis.

In this formula :

y tells us how far up the line goes
x tells us how far along
m is the Slope or Gradient i.e. how steep the line is
b is the Y-intercept i.e. where the line crosses the Y axis

The X and Y intercepts and the Slope are called the line properties. We shall now graph the line y-12x+7 = 0 and calculate its properties

According to: https://www.tiger-algebra.com/drill/y=12x-7/
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Jasmine knows that the area of a rectangle is the product of its base and height. Help her write an expression that represents t
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Answer:

Area of rectangle = H × B

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Step-by-step explanation:

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Height = H

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Area of rectangle

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WILL GIVE BRAINLIEST! SUPER CONFUSED!!!
kondor19780726 [428]

Answer:

For 1948 Men's :Interquartile range is 1.5, Median is 58.3

For 2012 Men's: Interquartile range is 0.315, Median is 47.86

You can infer that in 2012 the swimmers were better and it was more competitive as the interquartile range was lower and the median was also lower as well

Step-by-step explanation:

1948 Men's 100m

57.3, 57.8, 58.1, 58.3, 58.3, 59.3, 59.6,1:00.5

               ⬆                ⬆                 ⬆

         Low IQR      Median       High IQR

Low IQR is average of 57.8 and 58.1 = 57.95

High IQR  is average of 59.3 and 59.6 =  59.45

Median is average of 58.3 and 58.3 = 58.3

Interquartile range is High IQR- Low IQR

59.45 - 57.95=1.5

2012 Men's 100m

47.52, 47.53, 47.8, 47.84, 47.88, 47.92, 48.04, 48.44

                    ⬆                  ⬆                      ⬆

             Low IQR         Median          High IQR

Low IQR is average of 47.53 and 47.8 = 47.665

High IQR is average of 48.04 and 47.92 = 47.98

Median is average of 47.88 and 47.84 = 47.86

Interquartile range is High IQR-Low IQR

47.98-47.665=0.315

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3 years ago
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