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3 years ago

The cost of a car, in dollars, t years after its purchase is $20,000 - $3,000t. Which statement is correct? please explain too.

Mathematics

1 answer:

xeze [42]3 years ago

5 0

We start with 20,000. Every year t, we decrease 3,000. Thus, each year the car depreciates by 3,000 every year.

B

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A pyramid has a regular hexagonal base with side links of four and a slant height of six. Find the total area of the pyramid

vova2212 [387]

Area of the Base = 6 * side^2 / 4 * tan (180/6)

Area of the Base = 6 * 16 / 4 * 0.57735

Area of the Base = 96 / 2.3094

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Area of 6 Faces = 72

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3 years ago

1. Below is data collected from a random sample of 40 people at the public library. If the public library has 300 people, then w

nataly862011 [7]

Answer:

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2 years ago

Find the average value of the function on the given interval. 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

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8 0

2 years ago

Nico can do 42 push-ups in 3 minutes.

Alchen [17]

Step-by-step explanation:

Hey there!

It is said we need to find the unit rate.

What we can do is apply Unitary Method.

3 minutes = 42 push ups

1 minute = $\sf{\bf{\dfrac{42}{3}}}$

So, 12 push ups per minute is the unit rate.

In 5 minutes, he will do 12×5 = 60 push ups!

Hope it helps :)

4 0

2 years ago

See attached picture

yanalaym [24]

Answer:

$\frac{f(x+h)-f(x)}{h}$ $=2x + h$

Step-by-step explanation:

Given

$f(x)= x^2 + 2$

Required

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First, we calculate f(x + h)

$f(x)= x^2 + 2$

$f(x+h) = (x+h)^2+2$

$f(x+h) = x^2+2xh+h^2+2$

So, we have:

$\frac{f(x+h)-f(x)}{h}$ $= \frac{x^2 + 2xh + h^2 + 2 - x^2 - 2}{h}$

$\frac{f(x+h)-f(x)}{h}$ $= \frac{x^2 - x^2+ 2xh + h^2 + 2 - 2}{h}$

$\frac{f(x+h)-f(x)}{h}$ $= \frac{2xh + h^2}{h}$

$\frac{f(x+h)-f(x)}{h}$ $=2x + h$

8 0

2 years ago