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Alenkinab [10]
2 years ago
15

Can I have some help, I’ll mark you brainliest!!

Mathematics
2 answers:
riadik2000 [5.3K]2 years ago
6 0

Answer:

7

Step-by-step explanation:

NemiM [27]2 years ago
6 0
The anwser is 7 if it’s not I don’t know
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Susie said she could earl $600.00 if she sold 70 products and Max said he could earn $ 600.00 if he sold 90 products. Who is cor
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A calculator could help you out with problems like these...
8 0
3 years ago
If f(x)=-4x+4 then f^-1(x)=
PolarNik [594]

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f^-1=(-1/4)x+1

Step-by-step explanation:

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4 0
3 years ago
A model for the average price of a pound of white sugar in a certain country from August 1993 to August 2003 is given by the fun
Dovator [93]

Answer:

t = 0.811\,s contains the cheapest reference to sugar; t = 4.511\,s contains the most expensive reference to sugar.

Step-by-step explanation:

Let be s(t) = -0.00003237\cdot t^{5} + 0.0009037\cdot t^{4}-0.008956\cdot t^{3}+0.03629\cdot t^{2}-0.04547\cdot t + 0.4778, the times when sugar is the cheapest and the most expensive (absolute minimum and maximum) are determined with the help of first and second derivatives of this function (First and Second Derivative Tests):

First Derivative Test

s'(t) = -0.00016185\cdot t^{4}+0.0036148\cdot t^{3}-0.026868\cdot t^{2}+0.07258\cdot t - 0.04547

Let equalize the polynomial to zero and solve the resulting expression:

-0.00016185\cdot t^{4}+0.0036148\cdot t^{3}-0.026868\cdot t^{2}+0.07258\cdot t - 0.04547 = 0

t_{1} \approx 9.511\,s, t_{2}\approx 7.431\,s, t_{3}\approx 4.511\,s and t_{4}\approx 0.881\,s

Second Derivative Test

s''(t) = -0.0006474\cdot t^{3}+0.0108444\cdot t^{2}-0.053736\cdot t+0.07258

This function is now evaluated at each root found in the First Derivative section:

s''(9.511\,s) = -0.0006474\cdot (9.511\,s)^{3}+0.0108444\cdot (9.511\,s)^{2}-0.053736\cdot (9.511\,s)+0.07258

s''(9.511\,s) = -0.015 (A maximum)

s''(7.431\,s) = -0.0006474\cdot (7.431\,s)^{3}+0.0108444\cdot (7.431\,s)^{2}-0.053736\cdot (7.431\,s)+0.07258

s''(7.431\,s) = 6.440\times 10^{-3} (A minimum)

s''(4.511\,s) = -0.0006474\cdot (4.511\,s)^{3}+0.0108444\cdot (4.511\,s)^{2}-0.053736\cdot (4.511\,s)+0.07258

s''(4.511\,s) = -8.577\times 10^{-3} (A maximum)

s''(0.811\,s) = -0.0006474\cdot (0.811\,s)^{3}+0.0108444\cdot (0.811\,s)^{2}-0.053736\cdot (0.811\,s)+0.07258

s''(0.811\,s) = 0.036 (A minimum)

Each value is evaluated in order to determine when sugar was the cheapest and the most expensive:

Cheapest (Absolute minimum)

s(0.811\,s) = -0.00003237\cdot (0.811\,s)^{5}+0.0009037\cdot (0.811\,s)^{4}-0.008956\cdot (0.811\,s)^{3}+0.03629\cdot (0.811\,s)^{2}-0.04547\cdot (0.811\,s)+0.4778

s(0.811\,s) = 0.460

s(7.431\,s) = -0.00003237\cdot (7.431\,s)^{5}+0.0009037\cdot (7.431\,s)^{4}-0.008956\cdot (7.431\,s)^{3}+0.03629\cdot (7.431\,s)^{2}-0.04547\cdot (7.431\,s)+0.4778

s(7.431\,s) = 0.491

t = 0.811\,s contains the cheapest reference to sugar.

Most expensive (Absolute maximum)

s(4.511\,s) = -0.00003237\cdot (4.511\,s)^{5}+0.0009037\cdot (4.511\,s)^{4}-0.008956\cdot (4.511\,s)^{3}+0.03629\cdot (4.511\,s)^{2}-0.04547\cdot (4.511\,s)+0.4778

s(4.511\,s) = 0.503

s(9.511\,s) = -0.00003237\cdot (9.511\,s)^{5}+0.0009037\cdot (9.511\,s)^{4}-0.008956\cdot (9.511\,s)^{3}+0.03629\cdot (9.511\,s)^{2}-0.04547\cdot (9.511\,s)+0.4778

s(9.511\,s) = 0.498

t = 4.511\,s contains the most expensive reference to sugar.

4 0
3 years ago
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