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2 years ago

Consider functions f and g defined by the graph and the table. Please help me ASAP

Mathematics

1 answer:

Fed [463]2 years ago

3 0

Answer:

Step-by-step explanation:

Given are two functions represented one by graph and another by table

We can check the given statement one by one

1) $f(2)=18 : f(0) =15$

Average rate of change = $\frac{18-15}{2-0} =1.5$

For g(x) avg rate of change

= $\frac{g(2)-g(0)}{2-0} \\=4$

Hence g has a higher average rate of change than f.

True

2)f is linear and eventually exceeds g (f(0)>g(0) . True

3) True because y intercept is 10 for f while 8 for g

4) True because f(3) <g(3)

5) False both cannot be linear

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What is the area of the composite figure?

german

Answer:

16 + 6pi cm

Step-by-step explanation:

We have a square of length 4 cm

A = s^2 = 4^2 = 16

We have 3 semi circles

with radius 2

A semi circle has an area of

1/2 pi r^2 = 1/2 pi (2)^2 = 1/2 (4pi) = 2pi

There are 3 of them

3 * 2 pi = 6pi

Add the areas together for the square and the semicircles

16 + 6pi

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3 years ago

the acceleration of an object due to gravity is 32 feet per second squared.what is acceleration due to gravity in inches per sec

seropon [69]

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3 years ago

A 1970 comic book has appreciated 7% per year and originally sold for $0.35. What will the comic book be worth in 2017?

omeli [17]

You tell me if this makes sense to you.

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I hope this helps :)

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3 years ago

Charmaine is on the swim team. Each week she swims a total of 6,000 meters. How many kilometers does she swim each week?

alex41 [277]

Answer:

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3 years ago

Read 2 more answers

Find a particular solution to y" - y + y = 2 sin(3x)

leonid [27]

Answer with explanation:

The given differential equation is

y" -y'+y=2 sin 3x------(1)

Let, y'=z

y"=z'

$\frac{dy}{dx}=z\\\\d y=zdx\\\\y=z x$

Substituting the value of , y, y' and y" in equation (1)

z'-z+zx=2 sin 3 x

z'+z(x-1)=2 sin 3 x-----------(1)

This is a type of linear differential equation.

Integrating factor

$=e^{\int (x-1) dx}\\\\=e^{\frac{x^2}{2}-x}$

Multiplying both sides of equation (1) by integrating factor and integrating we get

$\rightarrow z\times e^{\frac{x^2}{2}-x}=\int 2 sin 3 x \times e^{\frac{x^2}{2}-x} dx=I$

$I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx -\int \frac{2\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx-\frac{2I}{3}\\\\\frac{5I}{3}=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{5}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{5} dx$

8 0

3 years ago