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3 years ago

5

If g(c)=9/5C+32 find g^-1(41)? Does anyone know the answer?

1 answer:

Scilla [17]3 years ago

7 0

C(f)=(5/9)(f-32) I think

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#9 i need help it would really be appreciated

son4ous [18]

For number 9, you know that it makes 40 copies every five minutes. All you have to do is divide 40 by 5.

40 ÷ 5 = 8

It makes 8 copies every minute.

Your chart can also help you check your answer. It says that it makes 96 copies every 12 minutes and it makes 104 copies every 13 minutes. So multiply 8 by 12 to check it and 8 by 13.

8 • 12 = 96
8 • 13 = 104

Since those answers match up with the chart, it must be correct.

The photocopier makes 8 copies every minute. Hope that helps!

4 0

3 years ago

What is the slope of the line shown in the graph? (4 points)<br> 5<br> 4<br> 1<br> 0<br> 10

Softa [21]

Answer:

There is no graph

Step-by-step explanation:

Post the graph and I can help you.

6 0

3 years ago

Select all of the expressions that are equivalent to -7.

Ivanshal [37]

Answer:

the answers that are equivalent are a, b, and e

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3 years ago

Use the intermediate value theorem to find the value of c such that f(c) = M. f(x) = x^2 - x + 1 text( on ) [1,8]; M = 21 c =

DanielleElmas [232]

Answer:

$c = 5$

Step-by-step explanation:

Given

$f(c) = M$

$f(x) = x^2 - x + 1$

$Interval: [1,8]$

$M = 21$

Required

Find c using Intermediate Value theorem

First, check if the value of M is within the given range:

$f(x) = x^2 - x + 1$

$f(1) = 1^2 - 1 + 1$

$f(1) = 1$

$f(x) = 8^2 - 8 + 1$

$f(x) = 57$

$1 \le M \le 57$

$1 \le 21 \le 57$

M is within range.

Solving further:

We have:

$f(c) = f(x) = M$

$f(x) = 21$

Substitute 21 for f(x) in $f(x) = x^2 - x + 1$

$21 = x^2 - x + 1$

Express as quadratic function

$x^2 - x + 1 - 21 = 0$

$x^2 - x - 20 = 0$

Expand

$x^2 + 4x - 5x - 20$

$x(x+4)-5(x+4)=0$

$(x - 5)(x+4) = 0$

$x - 5 = 0$ or $x + 4= 0$

$x = 5$ or $x = -4$

The value of $x = -4$ is outside the $Interval: [1,8]$

So:

$x = 5$

$f(c) = f(x) = M$

$f(c) = f(5) = 21$

By comparison:

$c = 5$

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3 years ago

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Ludmilka [50]

14 but i don't mind helping out anyone / everyone i can

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3 years ago

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