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gavmur [86]
4 years ago
14

Suppose that a population begins at a size of 100 and grows continuously at a rate of 200% per year. Give the formula for calcul

ating the size of that population after t years. Question 3 options: A) A = 100 + 2et B) A = 100 + te2 C) A = 100e2t D) A = 100 + e2t
Mathematics
2 answers:
mr Goodwill [35]4 years ago
7 0

Answer:

100 (+1) x 2^t     = 100 (+2et) as we plus and we multiply before squaring the years.

Step-by-step explanation:

You cna now work this out as +1 represents the 100% + 200% sum.

ankoles [38]4 years ago
3 0

Answer:

C) A = 100e^{2t}

Step-by-step explanation:

Continuous growth:

A = Pe^{rt}

r = rate as a decimal

r = 200% = 2

P = initial value = 100

A = 100e^{2t}

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stepan [7]
16(0.32) + 8(0.24) = c
5 0
3 years ago
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Suppose that the height (in centimeters) of a candle is a linear function of the amount of time (in hours) it has been burning.
kolbaska11 [484]

If the height of candle after 10 and 26 hours are 25 cm and 17 cm then the height of candle after 21 hours is 19.5 cm.

Given height of candle after 10 hours is 25 cm , height of candle after 26 hours is 17 cm.

We have to find the height of candle after 21 hours.

We have been given two points of linear function (10,25),(26,17).

We have to first form an equation which shows the height of candle after x hours.

let the hours be x and the height be y.

Equation from two points will be as under:

(y-y_{1} )=(y_{2} -y_{1} )/(x_{2} -x_{1} )*(x- x_{1} )

(y-25)=(17-25)/(26-10)* (x-10)

y-25=-8/16 *(x-10)

16(y-25)=-8(x-10)

16y-400=-8x+80

8x+16y=480

Now we have to put x=21 to find the height of candle after 21 years.

8*21+16y=480

168+16y=480

16y=480-168

16y=312

y=312/16

y=19.5

Hence the height of candle after 21 hours is 19.5 cm.

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6 0
2 years ago
4.63 rounded to the nearest tenth?
Tju [1.3M]
4.6 because you round up because the next number is less than 5
4 0
3 years ago
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What type of angles are these?
Stels [109]
Those angles are less than 90° so they would be acute angles
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4 years ago
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How do you find the solution area to a system of inequalities? 100 points
larisa [96]

How To Solve Systems of Inequalities Graphically

1) Write the inequality in slope-intercept form or in the form  

y

=

m

x

+

b

y=mx+b

.

For example, if asked to solve  

x

+

y

≤

10

x+y≤10

, we first re-write as  

y

≤

−

x

+

10

y≤−x+10

.

2) Temporarily exchange the given inequality symbol (in this case  

≤

≤

) for just equal symbol. In doing so, you can treat the inequality like an equation. BUT DO NOT forget to replace the equal symbol with the original inequality symbol at the END of the problem!

So,  

y

≤

−

x

+

10

y≤−x+10

becomes  

y

=

−

x

+

10

y=−x+10

for the moment.

3) Graph the line found in step 2. This will form the "boundary" of the inequality -- on one side of the line the condition will be true, on the other side it will not. Review how to graph a line here.

4) Revisit the inequality we found before as  

y

≤

−

x

+

10

y≤−x+10

. Notice that it is true when y is less than or equal to. In step 3 we plotted the line (the equal-to case), so now we need to account for the less-than case. Since y is less than a particular value on the low-side of the axis, we will shade the region below the line to indicate that the inequality is true for all points below the line:

 

5) Verify. Plug in a point not on the line, like (0,0). Verify that the inequality holds. In this case, that means  

0

≤

−

0

+

10

0≤−0+10

, which is clearly true. We have shaded the correct side of the line.

5 0
3 years ago
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