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3 years ago

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Use the functions h(x) = 2x + 5 and t(x) = 7x − 6 to complete the function operations listed below.

1 answer:

wlad13 [49]3 years ago

6 0

We have to use the functions:
h ( x ) = 2 x + 5 and t ( x ) = 7 x - 6
Part A:
( h + t ) ( x ) = ( 2 x + 5 ) + ( 7 x - 6 ) = 9 x - 1
Part B :
( h · t ) ( x ) = ( 2 x + 5 ) · ( 7 x - 6 ) = 14 x² - 12 x + 35 x - 30 =
= 14 x² + 23 x - 30
Part C :
h [ t ( x ) ] = h ( 7 x - 6 ) = 2 · ( 7 x - 6 ) + 5 = 14 x - 12 + 5 =
= 14 x - 7

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$\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}$

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4 years ago

The function in Exercise represents the rate of flow of money in dollars per year. Assume a 10-year period at 8% compounded cont

anzhelika [568]

Answer:

a) The present value is 688.64 $

b) The accumulated amount is 1532.60 $

Step-by-step explanation:

<u>a)</u><u> The preset value equation is given by this formula:</u>

$P=\int^{T}_{0}f(t)e^{-rt}dt$

where:

T is the period in years (T = 10 years)
r is the annual interest rate (r=0.08)

So we have:

$P=\int^{T}_{0}(0.01t+100)e^{-rt}dt$

Now we just need to solve this integral.

$P=\int^{T}_{0}0.01te^{-rt}dt+\int^{T}_{0}100e^{-rt}dt$

$P=e^{-0.08t}(-1.56-0.13t)|^{10}_{0}+1250e^{-0.08t}|^{10}_{0}$

$P=0.30+688.34=688.64 $$

The present value is 688.64 $

<u>b)</u><u> The accumulated amount of money flow formula is:</u>

$A=e^{r\tau}\int^{T}_{0}f(t)e^{-rt}dt$

We have the same equation but whit a term that depends of τ, in our case it is 10.

So we have:

$A=e^{r\tau}\int^{T}_{0}(0.01t+100)e^{-rt}dt=e^{0.08\cdot 10}P$

$A=e^{0.08\cdot 10}688.64=1532.60 $$

The accumulated amount is 1532.60 $

Have a nice day!

6 0

3 years ago