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3 years ago

9

Mary has gone into a department store with two coupons. One coupon is good for 20% off of the total not including tax. The other

coupon will take $15 off of her pre-tax total. If f(x) = 0.8x calculates the total after the 20% off coupon and g(x) = x – 15 calculates the total after the $15 dollar off coupon, determine g(f(100)). Then describe what g(f(100)) represents in the context of this problem.

1 answer:

lys-0071 [83]3 years ago

6 0

Answer:

The value of g(f(100)) is 65.

Step-by-step explanation:

It is given that one coupon is good for 20% off of the total not including tax. The other coupon will take $15 off of her pre-tax total.

The given functions are

$f(x)=0.8x$

$g(x)=x-15$

where, f(x) calculates the total after the 20% off coupon and g(x) calculates the total after the $15 dollar off coupon.

We need to find the value of g(f(100)).

$g(f(x))=g[0.8(x)]$ $(\because f(x)=0.8x)$

Substitute x=100 in the above function.

$g(f(100))=g[0.8(100)]$

$g(f(100))=g(80)$

Substitute x=80 in function g(x) to find the value of g(f(100)).

$g(f(100))=80-15$ $(\because g(x)=x-15)$

$g(f(100))=65$

Therefore, the value of g(f(100)) is 65.

g(f(x)) represents the value goods after applying both coupons consecutively.

Therefore, g(f(100)) represents the value of $100 goods after applying both coupons consecutive is 65.

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3 years ago

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Here the given points are (a, b) = (-4, 1) and (c, d) = (2, 4)

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3 years ago

A washer and a dryer cost 860 combined. The washer costs 90 less than the dryer. What is the cost of the dryer?

Kipish [7]

Answer:

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Step-by-step explanation:

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4 0

2 years ago

An easy way to explain...

Mashcka [7]

it's very simple, 694-72= 622
so in order for you to check you answer you do the opposite
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3 years ago

Read 2 more answers

The logistic equation for the population (in thousands) of a certain species is given by:

Eva8 [605]

Answer:

a.

b. 1.5

c. 1.5

d. No

Step-by-step explanation:

a. First, let's solve the differential equation:

$\frac{dp}{dt} =3p-2p^2$

Divide both sides by $3p-2p^2$ and multiply both sides by dt:

$\frac{dp}{3p-2p^2}=dt$

Integrate both sides:

$\int\ \frac{1}{3p-2p^2} dp =\int\ dt$

Evaluate the integrals and simplify:

$p(t)=\frac{3e^{3t} }{C_1+2e^{3t}}$

Where C1 is an arbitrary constant

I sketched the direction field using a computer software. You can see it in the picture that I attached you.

b. First let's find the constant C1 for the initial condition given:

$p(0)=3=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-1$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } =\frac{3}{2} =1.5$

c. As we did before, let's find the constant C1 for the initial condition given:

$p(0)=0.8=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=1.75$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2+1.75e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } =\frac{3}{2} =1.5$

d. To figure out that, we need to do the same procedure as we did before. So, let's find the constant C1 for the initial condition given:

$p(0)=2=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-\frac{1}{2} =-0.5$

Can a population of 2000 ever decline to 800? well, let's find the limit of the function when it approaches to ∞:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-0.5e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } =\frac{3}{2} =1.5$

Therefore, a population of 2000 never will decline to 800.

6 0

3 years ago