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3 years ago

Solve 6-8 and one and two

Mathematics

1 answer:

AfilCa [17]3 years ago

4 0

For 1, simply draw an equilateral triangle. The angles should be 60°, 60°, 60°. The sides can be any length, but they should become equal if you just draw out one 60° angle and connect the two edges.
For 2, just draw a right triangle with angles of 90° (obviously), and two others that are not 45° each. So, you could draw a 30°, 60°, 90° triangle, for example.

For 6-8, you have to just remember that when you add up the angle measures of a triangle, the total will always be 180°.

So, for 6, the answer is 60°.
For 7, the answer is 118°.
For 8, the answer is 27°.

Hope this helped!! xx

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Answer:

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number of tiles = 21.6/0.0576

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Find the average rate of change of f(x)=2x-1 from x=-1 to x=2

MrMuchimi

The average rate of change from x = -1 to x = 2 is 2

Solution:

Given function is:

f(x) = 2x - 1

We have to find the average rate of change from x = -1 to x = 2

The average rate of change is given as:

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

The average rate of change from x = -1 to x = 2 is given by formula:

$\text {Average rate of change}=\frac{f(2)-f(-1)}{2-(-1)}\\\\\text {Average rate of change}=\frac{f(2)-f(-1)}{3}$

Find f(2) and f( - 1)

Substitute x = 2 in given function

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Substitute x = -1 in given function

f( - 1) = 2(-1) - 1 = -2 - 1 = -3

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3 years ago

1. Contaminated water is subjected to a cleaning process. The concentration of the

Leokris [45]

Answer:

The concentration of pollutants in the water can be defined as the quotient between the mass of pollutants and the total mass of water.

Assuming that 1L of water weighs 1000 grams, we know that:

Mass of pollutants = 5mg = 0.005g.

Initial concentration = 0.005g/1000g = 0.000005

If we multiply this by 100%, we get the percentage:

0.000005*100% = 0.0005%

Now, the mass of pollutants decreases by 10% each hour.

So if initially, we have 5mg

After one hour, we will have: 5mg - 0.1*5mg = 5mg*(0.9).

After another hour, we will have: 5mg*(0.9) - 0.1*5mg*0.9 = 5mg*(0.9)^2.

And so on, then after n hours, the mass of pollutants will be:

M(n) = 5mg*(0.9)^n

Or we can write this in grams as:

M(n) = 0.005g*(0.9)^n

Then the concentration as a function of time in hours will be:

C(n) = M(n)/1000g = (0.005g/1000g)*(0.9)^n

Notice that the thing between the parentheses is the initial concentration, then if we write this in percentage form:

c(n) = 100%*(0.005g/1000g)*(0.9)^n = 0.0005%*(0.9)

The function that represents the concentration of polution in the water as a function of hours is:

c(n) = 0.0005%*(0.9)

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