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3 years ago

Timmy has 13 dollars in his bank account he withdrew m amount of dollars. He is now 100 dallors in debt. How much

Mathematics

2 answers:

Gekata [30.6K]3 years ago

8 0

Answer:

$113

Step-by-step explanation:

Since he had $13 and now he is $100 in debt:

First, he withdrew $13 to get $0 in his bank account.

He then withdrew $100 to get -$100 in his bank account.

13 + 100 = 113

enyata [817]3 years ago

3 0

Answer:

Timmy withdrew $113

Step-by-step explanation:

The following equation represents that situation:

$13 - m = -$100

Adding m to both sides, we get:

$13 = m - $100

Combining like thers, we get m = $113

Timmy withdrew $113

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Trevor is making a measuring strip. He says, I don’t want to leave gaps so I made sure each ten strip was a little bit on top of

Romashka [77]

Answer:

the measurements will be wrong because they are overlapping each other. he needs to line them up correctly and measure that way

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4 years ago

41 tens×4hundreds=<br> 410×400=

weqwewe [10]

410x400 is correct and your answer would be 164,000.

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3 years ago

On a coordinate plane, a line is drawn from point A to point B. Point A is at (negative 4, 8) and point B is at (2, negative 4).

xxMikexx [17]

The x- and y-coordinates of point C, which partitions the directed line segment from A to B into the ratio 3:10 is; (-2.6, 5.2)

<h3>How to find coordinate points?</h3>

The coordinate of the points are given as:

A(-4, 8)

B(2, -4)

Ratio of partition is given as; 3:10

Now, the point at the given ratio is calculated from the formula;

C = [(mx2 + nx1)/(m + n)], [(my2 + ny1)/(m + n)]

Thus;

C = [(3*2 + 10*-4)/(3 + 10)], [(3*-4 + 10*8)/(3 + 2)]

C = (-34/13, 68/13)

C = (-2.6, 5.2)

Read more about Coordinate Points at; brainly.com/question/17206319

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6 0

2 years ago

John has an annual salary of $48,000.00 and he is paid every two weeks. What will his

yanalaym [24]

Answer: $1846

Step-by-step explanation:

First and foremost, we should note that there are 52 weeks in a year. Therefore the biweekly payment of John will be gotten by dividing his gross income by 26 weeks i.e (52weeks/2) since it's biweekly.

Therefore, the gross income on each paycheck will be:

= $48,000.00 / 26

= $1846.15

= $1846 approximately

The gross income on each paycheck is $1846.

8 0

3 years ago

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