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3 years ago

Application of Linear Function

Mathematics

1 answer:

Yakvenalex [24]3 years ago

4 0

Answer:

see below

Step-by-step explanation:

Find the slope

m= ( 40.12 - 11.02)/ (400-100)

= 29.1 / 300

=.097

We can use the linear form y = mx+b

y =.097x +b

40.12 = .097 ( 400) +b

40.12 = 38.80 +b

1.32 = b

y = 0.097x + b

C(x) = .097x + 1.32

The fixed cost is b = 1.32

C(1001) = .097(1001) + 1.32

=98.417

The marginal cost is the cost of adding one more item, which is the slope

.097 is the marginal cost

It means to make one more cup of coffee will cost $.097

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Papessa [141]

Answer:

Step-by-step explanation:

3*2x=6x, 3*5=10

So, 6x+10+4x = 5x+15

Add 6 and 4 which leaves you with 10x+10 = 5x+15

subtract 10 from 15 and that leaves you with 5 divided by 10x

x=2

Same for second equation

2*8x=16x and 2*3= 6

So 16x-6=16x+28

Add 6 to both sides so you get 34.

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3 years ago

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4 years ago

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Please explain and solve 2/3 = m/42

ivanzaharov [21]

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3 years ago

If person A earns $51000 more than person b. if total salaries are $304250. find salaries of a nd b

Dafna11 [192]

X + y = 304,250
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Plug in (y+51,000) for x into the first equation

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4 years ago

Combine into a single logarithm. 3log(x+y)+2log(x-y)-log(x^2 +y^2)

seropon [69]

Answer:

$3\log _{10}\left(x+y\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

Step-by-step explanation:

Given the expression

$3log\left(x+y\right)+2log\left(x-y\right)-log\left(x^2\:+y^2\right)$

solving to write into a single logarithm

$3log\left(x+y\right)+2log\left(x-y\right)-log\left(x^2\:+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)$

$3\log _{10}\left(x+y\right)=\log _{10}\left(\left(x+y\right)^3\right)$

$=\log _{10}\left(\left(x+y\right)^3\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)$

$2\log _{10}\left(x-y\right)=\log _{10}\left(\left(x-y\right)^2\right)$

$=\log _{10}\left(\left(x+y\right)^3\right)+\log _{10}\left(\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)$

$\log _{10}\left(\left(x+y\right)^3\right)+\log _{10}\left(\left(x-y\right)^2\right)=\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)$

$=\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)$

$\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

$=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

Thus,

$3\log _{10}\left(x+y\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

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3 years ago