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Goshia [24]
3 years ago
8

The graph below represents a population

Mathematics
1 answer:
Nataliya [291]3 years ago
6 0

Answer:

Average rate of  change for the function for the interval (6, 12] is 500 people per year.

Option A is correct.

Step-by-step explanation:

We need to find the average rate of  change for the function for the interval

(6, 12]

The formula used to calculate Average rate of change is:

Average \ rate \ of \ change=\frac{f(b)-f(a)}{b-a}

We are given a=6 and b=12

Looking at the graph we can see that when x=6 y= 3000 so, f(a)=3000

and when x=12, y=6000 so, f(b)=6000

Putting values in formula and finding Average rate of change:

Average \ rate \ of \ change=\frac{f(b)-f(a)}{b-a}\\Average \ rate \ of \ change=\frac{6000-3000}{12-6}\\Average \ rate \ of \ change=\frac{3000}{6}\\Average \ rate \ of \ change=500

So, average rate of  change for the function for the interval (6, 12] is 500 people per year.

Option A is correct.

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Answer:

\triangle EDF is isosceles.

Step-by-step explanation:

Please have a look at the attached figure.

We are <u>given</u> the following things:

\angle EDF = y

\text{External }\angle DFG = 90 +\dfrac{y}{2}

Let us try to find out \angle E and \angle DFE. After that we will compare them.

<u>Finding </u>\angle DFE<u>:</u>

Side EG is a straight line so \angle GFE = 180

\angle GFE is sum of internal \angle DFE and external \angle DFG

\angle GFE = 180 = \angle DFE  + \angle DFG\\\Rightarrow 180 = \angle DFE + (90+\dfrac{y}{2})\\\Rightarrow \angle DFE = 180 - 90 - \dfrac{y}{2}\\\Rightarrow \angle DFE = 90 - \dfrac{y}{2} ....... (1)

<u>Finding </u>\angle E<u>:</u>

<u>Property of external angle:</u> External angle in a triangle is equal to the sum of two opposite internal angles of a triangle.

i.e. external \angle DFG = \angle E + \angle EDF

\Rightarrow 90+\dfrac{y}{2} = \angle E + y\\\Rightarrow \angle E = 90+\dfrac{y}{2}  -y\\\Rightarrow \angle E = 90-\dfrac{y}{2} ....... (2)

Comparing equations (1) and (2):

It can be clearly seen that:

\angle DFE = \angle E =90-\dfrac{y}{2}

The two angles of \triangle EDF are equal hence \triangle EDF is isosceles.

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