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3 years ago

Whats the value of x?

Mathematics

1 answer:

Harrizon [31]3 years ago

6 0

Answer:

x=8.8125

Step-by-step explanation:

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Find the average value of the function on the given interval.<br> f(x)=2 ln x; [1,e]

JulijaS [17]

Answer:

$f_{avg}=\frac{1}{e-1}$

Step-by-step explanation:

We are given that a function

$f(x)=2lnx$

We have to find the average value of function on the given interval [1,e]

Average value of function on interval [a,b] is given by

$\frac{1}{b-a}\int_{a}^{b}f(x)dx$

Using the formula

$f_{avg}=\frac{1}{e-1}\int_{1}^{e}lnx dx$

By Parts integration formula

$\int(uv)dx=u\int vdx-\int(\frac{du}{dx}\int vdx)dx$

u=ln x and v=dx

Apply by parts integration

$f_{avg}=\frac{1}{e-1}([xlnx]^{e}_{1}-\int_{1}^{e}(\frac{1}{x}\times xdx))$

$f_{avg}=\frac{1}{e-1}(elne-ln1-[x]^{e}_{1})$

$f_{avg}=\frac{1}{e-1}(e-0-e+1)=\frac{1}{e-1}$

By using property lne=1,ln 1=0

$f_{avg}=\frac{1}{e-1}$

8 0

3 years ago

Find the selling price of each item.<br> 57) Original price of a puppy: 339.50<br> Discount: 55

densk [106]

Answer:

The selling price of the puppy is $284.50.

Step-by-step explanation:

In the present case, the puppy's original selling price is $ 339.50. At this price a discount of $ 55 is made, that is, that amount is subtracted from the original price, modifying the sale price.

Therefore, since 339.50 - 55 results in 284.50, the puppy's sale price goes from $ 339.50 to $ 284.50, taking the aforementioned discount.

5 0

3 years ago

An investment of $8,000 earns interest at an annual rate of 7% compounded continuously. Complete parts (A) and (B) below. Click

Sergeeva-Olga [200]

Answer:

A. $\mathtt{\dfrac{dA}{dt}|_{t=2}=644.15}$

B. $\mathtt{\dfrac{dA}{dt}|_{t = 5.79}= 839.86 }$

Step-by-step explanation:

Given that:

An investment of Amount = $8000

earns at an annual rate of interest = 7% = 0.07 compounded continuously

The objective is to :

A) Find the instantaneous rate of change of the amount in the account after 2 year(s).

we all know that:

$A = Pe^{rt}$

where;

$A = (8000) \ e ^{0.7t}$

The instantaneous rate of change = $\dfrac{dA}{dt}$

$\dfrac{dA}{dt} = \dfrac{d}{dt}(8000 \ e ^{0.07t} )$

$= 8000 \dfrac{d}{dt}e^{0.07 \ t}$

$\dfrac{dA}{dt}= 8000 (0.07)e^{0.07 \ t}$

$\dfrac{dA}{dt}= 560 e^{0.07 \ t}$

At t = 2 years; the instantaneous rate of change is:

$\dfrac{dA}{dt}|_{t=2}= 560 e^{0.07 \times 2}$

$\mathtt{\dfrac{dA}{dt}|_{t=2}=644.15}$

(B) Find the instantaneous rate of change of the amount in the account at the time the amount is equal to $12,000.

Here the amount = 12000

$12000 = (8000)e^{0.07 \ t}$

$\dfrac{12000 }{8000}= e^{0.07 \ t}$

$1.5= e^{0.07 \ t}$

㏑(1.5) = 0.07 t

0.405465 = 0.07 t

t = 0.405465 /0.07

t = 5.79

$\dfrac{dA}{dt}= 560 e^{0.07 \ t}$

At t = 5.79

$\dfrac{dA}{dt}|_{t = 5.79}= 560 e^{0.07 \times 5.79}$

$\mathtt{\dfrac{dA}{dt}|_{t = 5.79}= 839.86 }$

7 0

3 years ago

Which describes a person designated to receive money from a life insurance

Rainbow [258]

Answer: C

Step-by-step explanation: Beneficiary

8 0

3 years ago

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What is -3 2/3 times (-2 1/2)? A. -8 1/4 B. -6 1/6 C. 8 1/4 D. 8 3/4

Ivenika [448]

The answer would be c I think

3 0

3 years ago

Read 2 more answers