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Zarrin [17]
2 years ago
9

Help please!!! I don’t understand

Mathematics
1 answer:
Murljashka [212]2 years ago
3 0
I believe the answer is vertical stretch
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A client asks your catering company to provide a quote on a three-course dinner for 100 people. You know that it will cost you $
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For the function f(x)=(x+6)^3, rind f^-1(x)
qaws [65]

Answer:

f^{-1}(x)=\sqrt[3]{x}-6

Step-by-step explanation:

f(x)=(x+6)^3

y=(x+6)^3

x=(y+6)^3

\sqrt[3]{x}=y+6

\sqrt[3]{x}-6=y

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100 POINT QUESTION NEED HELP
amm1812

Answer:

The answer would be C

Step-by-step explanation:

the distance would be how far the two domains are from each other, therefore they are both added together

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2 years ago
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Nina and Jo both ran an 8 km race. Nina took 55 minutes to run the whole race. Jo started the race 3 minutes later than Nina but
grin007 [14]

Answer:

Step-by-step explanation:

First we figure out how fast Nina can run. If Nina can run 8 km in 55 minutes, then her rate is

\frac{8km}{55min}=.145\frac{km}{min} and we can use that in a d = rt table:

                 d        =        r        *        t

Nina                            .145

Jo

Now we can fill in the distance which is 6 for both, since that is the distance where they met:

                d        =        r       *        t

Nina         6        =     .145

Jo             6        =

Now we go to the info given about the time. If Jo started the race 3 minutes after Nina, that means that Nina is running 3 minutes longer than Jo. Filling in the time info:

                d        =        r        *        t

Nina          6       =       .145    *      t + 3

Jo              6       =         r       *         t

As you can see, right now we have 2 unknowns in Jo's row. But we don't have to! We will go to Nina's row where the only unknown is time and solve for t. If d = rt, then

6 = .145(t + 3) and

6 = .145t + .435 and

5.55 = .145t so

t = 38.379 minutes. This means that Jo was running 38.379 minutes when she caught up to Nina (it took Nina 3 minutes longer than that to go 6 km since she was already running for 3 minutes when Jo started the race). If Jo's time is 38.379, we can use that in her row for t and solve for r. If d = rt, then

6 = r(38.379) and

r = .16 km/min

Let's check it without the rounding (rounding takes away from the accuracy). If 6 = .145(t + 3) and Nina's rate not rounded is .145454545 and t = 38.37931034, then, rewriting without rounding:

6 should equal .145454545( 38.37931034 + 3)

6 ?=? .145454545(41.37931034)

6 ?=? 6.0 so

Jo's rate is .16 km/min rounded

6 0
2 years ago
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