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Morgarella [4.7K]

3 years ago

9

Two functions are shown in the table below: Function 1 2 3 4 5 6 f(x) = −x2 + 4x + 12 g(x) = x + 2 Complete the table on your ow

n paper, then select the value that is a solution to f(x) = g(x). x = 2 x = 3 x = 5 x = 6

1 answer:

zubka84 [21]3 years ago

3 0

Answer:

x f(x) g(x)

1 15 3

2 16 4

3 15 5

4 12 6

5 7 7

6 0 8

f(x) = g(x) when x = 5

Step-by-step explanation:

f(x) = -x² + 4x + 12

f(1) = -(1)² + 4(1) + 12 = 15

f(2) = -(2)² + 4(2) + 12 = 16

f(3) = -(3)² + 4(3) + 12 = 15

f(4) = -(4)² + 4(4) + 12 = 12

f(5) = -(5)² + 4(5) + 12 = 7

f(6) = -(6)² + 4(6) + 12 = 0

g(x) = x + 2

g(1) = 1 + 2 = 3

g(2) = 2 + 2 = 4

g(3) = 3 + 2 = 5

g(4) = 4 + 2 = 6

g(5) = 5 + 2 = 7

g(6) = 6 + 2 = 8

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Click here please. The problems are in picture format

padilas [110]

First picture:

You got the first two right (good job!), so I'll just tackle the third one. If you have to compute $f(x-4)$ , it means that you have to substitute $x-4$ in place of $x$ . Differently from the first two points, this will generate a new function, rather than a specific value:

$f(x) = \sqrt{x+4}+2 \implies f(x-4) = \sqrt{(x-4)+4}+2 = \sqrt{x}+2$

Second picture:

Given the point $(x,y)$ highlighted in the picture, you can deduce that the base is $2x$ units long (since it spans from $(-x,0)$ to $(x,0)$ ) and the height is $y$ units long (because it spans from $(x,0)$ to $(x,y)$ ). So, the area of the rectangle is the multiplication between base and height:

$A = 2xy$

But we know that $y=f(x)=\sqrt{36-x^2}$ , so we have

$A = 2x\sqrt{36-x^2}$

The domain of this function is given by the domain of the square root: we want its argument to be non, negative, so we have

$36-x^2 \geq 0 \iff x^2 \leq 36 \iff -6 \leq x \leq 6$

But since the problem is symmetric, the answer is

$0 \leq x \leq 6$

You can only see the answer $0 < x < 6$ because, if you choose $x=0$ or $x=6$ , the rectangle degenerates to a segment, and your exercise doesn't like this scenario, apparently

4 0

4 years ago

Please help I will mark brainliest

maksim [4K]

Answer:

<4 and <12

Step-by-step explanation:

When looking at Corresponding Angles, they need to be on opposite sides of a transversal. In this case, the answer is <4 and <12

7 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

Craig saved $10,950 in a year. He saved $75 more than what he saved the previous

Dmitry [639]

Answer: $10,875

Step 1: Subtract.

10,950 - 75 = 10,875.

Step 2: Add a dollar sign.

$10,875

Hope this helps!

7 0

3 years ago

Read 2 more answers

KIM [24]

A coordinate plane with the x-axis labeled Blueberries (in pints) and the y-axis labeled Cost (in dollars) shows a line going through (0, 0) and (5, 17).

<h3>What is directly proportional?</h3>

It is defined as the relationship between two quantities as one quantity increases the other quantity also increases and vice versa.

It is given that there is a proportional relationship between cost and pints of blueberries as,

Blueberries (in pints) 2 5

Cost (in dollars) 6.80 17

17/6.80 = 5/2

As we draw the graph and analyze each option one by one we get the graph shown by the statement C following the given proportional relationship.

Thus, a coordinate plane with the x-axis labeled Blueberries (in pints) and the y-axis labeled Cost (in dollars) shows a line going through (0, 0) and (5, 17).

Learn more about the directly proportional here:

brainly.com/question/16937826

#SPJ2

4 0

1 year ago