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3 years ago

If f(x)=2x+3 and g(x)=x^2, then g[f(-3)]=?

Mathematics

1 answer:

Wittaler [7]3 years ago

8 0

If f(x)=2x+3 and g(x)=x^2 then

f(g(x))=2x^2+3 then:

f(g(-3))=2(-3)^2+3

f(g(-3))=2*9+3

f(g(-3))=18+3

f(g(-3))=21

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Circle towns limit form a perfectly circular shape that has a population of 20000 in the population density of 480 people per sq

Alex

Answer:

$D =\frac{P}{\pi X^2}$

And solving for the radius we got:

$X = \sqrt{\frac{P}{\pi D}}$

And replacing the data given we got:

$X = \sqrt{\frac{480 \frac{people}{km^2}}{\pi *20000 people}}= 0.0874Km$

And this value converted to meters is $X = 87.40 m$

Step-by-step explanation:

For this case we know the population size $P = 20000$ and we also know the population density $D = 480 \frac{people}{km^2}$

We can assume that the area is a circle. We also know that the formula for the population density is given by:

$D= \frac{P}{A}$

Where P represent the number of people and A the area. Since we are assuming a circle then the area is given by:

$A = \pi X^2$

With X the radius of the circle

And then the populationd density become:

$D =\frac{P}{\pi X^2}$

And solving for the radius we got:

$X = \sqrt{\frac{P}{\pi D}}$

And replacing the data given we got:

$X = \sqrt{\frac{480 \frac{people}{km^2}}{\pi *20000 people}}= 0.0874Km$

And this value converted to meters is $X = 87.40 m$

6 0

3 years ago

If a cross section of the paperweight is cut perpendicular to the base but does not pass through the top vertex, which shape des

Montano1993 [528]

Rectangle shape paper weight

4 0

3 years ago

Please answer the question above.

Crank

Answer:

Waffle house :3

Step-by-step explanation:

8 0

3 years ago

I need help with this please

Doss [256]

Answer:I feel bad

Step-by-step explanation:

3 0

3 years ago

2. The number of atoms of a particular chemical element is modelled by

____ [38]

9514 1404 393

Answer:

a) 600

b) see below

c) 1.26 hours

Step-by-step explanation:

a) The value of y when x=0 is the coefficient of the exponential term:

y = 600·3^(-0) = 600·1 = 600

There were 600 atoms to start.

b) see attached for a graph

c) The graph shows 150 atoms at t = 1.26, about 1.26 hours after the start of time counting.

If you want to find that value algebraically, substitute for y and solve for x. Logarithms are involved.

150 = 600·3^(-x)

150/600 = 3^(-x)

log(1/4) = -x·log(3)

x = -log(1/4)/log(3) = log(4)/log(3) ≈ 1.2618595

After about 1.26 hours, there were 150 atoms.

6 0

3 years ago