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oksano4ka [1.4K]
3 years ago
15

HELP ASAP PLEASE i’ll give u brainliest

Mathematics
2 answers:
AnnZ [28]3 years ago
4 0

Answer:

1st Week-60

2nd Week-40

3rd Week-100

x=Amount During Second Week

(x)+(x+20)+(x+60)=200

3x+80=200

3x=120

x=40

1st Week=x+20=60

2nd Week=x=40

3rd Week=x+60=100 :)

Papessa [141]3 years ago
3 0

Answer:

During 3rd week, she earned $100, during second week she earned $40 and during 1st week she earned $60.

Step-by-step explanation:

Let money she earned in 1st week be x, 2nd week be y and 3rd week be z.

z = 60+y

z = 40+ x

now,

60+y = 40+ x

or, y = 40+x-60

or, y = x-20

Now,

x + y + z = 200

or, x + x - 20 + 40 + x = 200

or, 3x + 20 = 200

or, 3x = 180

so, x = $60

Now,

y = x-20

or, y = 60-20

so, y = $40

Now,

z = 60+y

or, z = 60+40

so, z = $100

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If f(x) = 245<br>Find f^-1(x) (the inverse) ​
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9514 1404 393

Answer:

  does not exist

Step-by-step explanation:

The notation f^-1(x) usually refers to the inverse function. That is, if we have the functional relation

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In order for an inverse function to exist, the original function must pass the "horizontal line test." That is, any horizontal line can intersect the function's graph in, at most, one point.

Here, the function f(x) = 245 is a horizontal line, so the horizontal line test fails. The horizontal line at y=245 will intersect f(x) in an infinite number of points.

So, f(x) = 245 does not have an inverse function. f^-1(x) does not exist.

_____

<em>Alternate interpretation of the question</em>

Sometimes, an exponent is applied to a function name when the intended operation is application of the exponent to the function value. For example, you may see sin²(x) when the intended meaning is sin(x)². The application of an exponent to the function name could be confused with multiple applications of the function: sin²(x) = sin(sin(x)), as the notation is sometimes used for this purpose, too.

If what you intend by f^-1(x) is really f(x)^-1, then you can say ...

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