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2 years ago

Consider the linear equation below.

Mathematics

1 answer:

densk [106]2 years ago

8 0

4- distributive property

Reason
X(y+u)= xy+xu
Expansion is distributive property

Hope it helps

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There are 140 people in a singing competition. The graph shows the results for the first five rounds

SOVA2 [1]

a) The expression is y = 140 - 15 x

b ) There are 5 participants in the 9th round

Step-by-step explanation:

Step 1 :

From the graph we can see that,

Number of people in the singing competition = 140

Number of people in the first round = 125

Number of people in the second round = 110

Number of people in the third round = 95

Number of people in the fourth round = 80

Number of people in the fifth round = 65

Step 2 :

From the given data, We can get that the difference between the number of people participating in each round is 15 less than the previous round .

The first round has 15 people less than the total number 140

Let x represent the number of the round and y represent the number of people participating in each round .

Then the expression to represent this would be

y = 140 - 15 x

Step 3 :

To find the number of participants in the 9th round given the same pattern continues.

For the 9th round x = 9, as x represents the number of the round

Substituting this in the equation obtained in step 2, we get

y = 140 -15 (9) = 140 - 135 = 5

There are 5 participants in the 9th round

Step 4 :

Answer :

a) The expression is y = 140 - 15 x

b ) There are 5 participants in the 9th round

4 0

4 years ago

What is the answer?? I need help!!<br><br> NO LINKS!!!

Charra [1.4K]

Function

Use "vertical line test"

searched it if you don't know

8 0

3 years ago

Read 2 more answers

Can you pleaseee help meee

Mrac [35]

Yesssssssss, with what though

6 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

The ratio of boys to girls in a group is 7:1. If there are 48 more boys than girls, work out how many girls there are.

Anvisha [2.4K]

Answer: 8 girls

Step-by-step explanation:

Difference between boys and girls = 7 - 1

= 6 units

6 units = 48

1 unit (girls) = 8

<em>I hope this helped! :)</em>

7 0

4 years ago