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3 years ago

Anyone good at this?? Please help me. I attempted at this and I don’t think my answer is correct!

Mathematics

1 answer:

AnnZ [28]3 years ago

3 0

Answer:

Step-by-step explanation:

There are two ways you could do this question. You could graph it with something like Desmos or you could put a couple of x values in for f(x). I'll do the latter first. When you understand it, try putting it into Desmos.

f(x) = (3/4)^x

f(0) = (3/4)^0

f(0) = 1

=======

f(x) = (3/4)^x

f(1) = 3/4

=========

g(x) = (3/4)^x - 6

g(0) = (3/4)^0 - 6

g(0) = 1 - 6

g(0) = - 5

========

g(x) = (3/4)^x - 6

g(1) = (3/4)^1 - 6

g(1)= 3/4 - 6

g(1) = -5 1/4

Conclusion

It looks like the graph has shifted down 6.

Answer

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3 years ago

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Malachai is doing a 20 miles bike ride. He has cycled 11 miles so far. What percentage of the the total distance has he cycled

Anestetic [448]

11 miles / 20 miles =.55
.55 x 100 = 55%
11/20 is the proportion and you multiply by 100 to gey a percent
another way to solve is
11/20 x 5/5 = 55/100 = 55%
you multiply by 5/5 because it is equal to one and it makes the denominator equal to 100 it also follows the rule that whatever you do to the top you must do with the bottom

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3 years ago

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erma4kov [3.2K]

Answer:

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Step-by-step explanation:

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You were paid $82.50 for seven 1/2 hours <br> of work what is your rate of pay

sertanlavr [38]

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3 years ago

Will someone help with this ! I was allowed a retry for it and I don’t know how to do it

pashok25 [27]

Answer:

$\frac{dN}{dt}=k(725-N)\\separating the variables and integrating\\\int \frac{dN}{725-N}=\int kdt+c\\-log (725-N)=kt+c\\log (725-N)=-kt-c\\(725-N)=e^{-kt-c} =e^{-kt} *e^{-c} =Ce^{-kt} \\when t=0,N=400\\725-400=Ce^{0} =C\\C=325\\when t=3,N=650\\725-650=325e^{-3k} \\\frac{75}{325}=(e^{-k})^3\\\\e^{-k} =(\frac{3}{13}) ^{\frac{1}{3} } \\when t=5\\725-N=325(\frac{3}{13}) ^{\frac{5}{3} } =325*\frac{3}{13}*(\frac{3}{13} )^{\frac{2}{3}}\\N=725-75*(\frac{3}{13} )^{\frac{2}{3} }=725-28=697$

Step-by-step explanation:

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3 years ago