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Kay [80]
3 years ago
11

Give m an example of age problem

Mathematics
2 answers:
Anit [1.1K]3 years ago
6 0
Q: Kevin is 4 years older than Margaret. 
Next year Kevin will be 2 times as old as Margaret.
<span>How old is Kevin? 
A:</span>
Solution 
<span>Denote as </span><span> Kevin's present age.</span>
<span>Then Margret's present age is </span><span>. </span>

<span>Next year Kevin will be </span><span> years old, and Margaret will be </span><span> years old. </span>

<span>Since next year Kevin will be 2 times as old as Margaret, you can write the equation </span>
<span>. </span>

<span>Solve this equation by simplifying it, step by step: </span>

<span>   (after brackets opening at the right side)</span>
<span>   (after moving variable terms to the right and constant terms to the left)</span>
<span>             (after combining like terms) </span>

<span>Thus, you got that Kevin's present age is </span><span> years. </span><span>


</span>
Finger [1]3 years ago
6 0
Jenifer birthday is today she will turn into 4 years if she collect 23 gifts and if there is 1 Dozen if candy ( dozen mean 12) how many candy does she have in all. (A) ( 12 candy) (b) (276 candy) (c) (35 candy) (d) (23 candy) (Hint multiply the number that's given on problem) it's easy as abc if u know your times table solve it
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Calculate a rough estimate of the total lineal footage of lumber needed for a rectangular house 35’ x 40’?
Maru [420]

Answer:

B. 1,500’

Step-by-step explanation:

Given;

dimension of the rectangular house, = 35’ x 40’

The actual total linear footage of lumber needed = 35’ x 40’ = 1,400'

The rough estimate should be greater than the actual estimate in 100s;

rough estimate = 1,400' + 100' = 1,500'

Therefore, the rough estimate of the total linear footage of lumber needed is 1,500'.

7 0
3 years ago
Draw a diagram to show $342 in $100 bills, $10 bills, and $1 bills
MArishka [77]
3 100 hundred bills 4 ten dollar bills and 2 one dollar bills
3 0
3 years ago
You buy 3.18 pounds of ​apples, 1.35 pounds of oranges​, and 2.25 pounds of pears. What is your total​ bill?
kykrilka [37]

Answer:

6.78

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Population Growth A lake is stocked with 500 fish, and their population increases according to the logistic curve where t is mea
Alexus [3.1K]

Answer:

a) Figure attached

b) For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

c) p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}

And if we find the derivate when t=1 we got this:

p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114

And if we replace t=10 we got:

p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404

d) 0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And then:

0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)

0 =19e^{-\frac{t}{5}} -1

ln(\frac{1}{19}) = -\frac{t}{5}

t = -5 ln (\frac{1}{19}) =14.722

Step-by-step explanation:

Assuming this complete problem: "A lake is stocked with 500 fish, and the population increases according to the logistic curve p(t) = 10000 / 1 + 19e^-t/5 where t is measured in months. (a) Use a graphing utility to graph the function. (b) What is the limiting size of the fish population? (c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months? (d) After how many months is the population increasing most rapidly?"

Solution to the problem

We have the following function

P(t)=\frac{10000}{1 +19e^{-\frac{t}{5}}}

(a) Use a graphing utility to graph the function.

If we use desmos we got the figure attached.

(b) What is the limiting size of the fish population?

For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

(c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months?

For this case we need to calculate the derivate of the function. And we need to use the derivate of a quotient and we got this:

p'(t) = \frac{0 - 10000 *(-\frac{19}{5}) e^{-\frac{t}{5}}}{(1+e^{-\frac{t}{5}})^2}

And if we simplify we got this:

p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}

And if we simplify we got:

p'(t) =\frac{38000 e^{-\frac{t}{5}}}{(1+19e^{-\frac{t}{5}})^2}

And if we find the derivate when t=1 we got this:

p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114

And if we replace t=10 we got:

p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404

(d) After how many months is the population increasing most rapidly?

For this case we need to find the second derivate, set equal to 0 and then solve for t. The second derivate is given by:

p''(t) = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And if we set equal to 0 we got:

0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And then:

0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)

0 =19e^{-\frac{t}{5}} -1

ln(\frac{1}{19}) = -\frac{t}{5}

t = -5 ln (\frac{1}{19}) =14.722

7 0
3 years ago
The table below shows data on the number of emergency room admissions recorded over the span of one shift.
densk [106]

Answer:

C. III

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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