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3 years ago

This is the full page plz help

Mathematics

1 answer:

Scorpion4ik [409]3 years ago

3 0

Angle 1 is congruent to angles 3, 5, and/or 7

Angle 2 is congruent to angles 4, 6, and/or 8

Angle 5 is congruent to angles 7, 3 and/or 1

Angle 6 is congruent to angles 8, 4, and/or 2

Any of these answers could work for the blanks.

Angles 1 and 3, 2 and 4, 5 and 7, and angles 6 and 8 are congruent because they are vertical angles. They have the same vertex. Not all of these are congruent to each other if this doesn’t make sense. It’s only 1 is congruent to 3, 2 congruent to 4, etc.

Then you have your corresponding angles. These are ones like angles 2 and 6, then 1 and 5. You can also have 8 and 4, or 7 and 3 as corresponding angles

Transversal angles are different. This would be like angles 3 and 4, or 1 and 2. They are not always congruent. The only time they will be congruent is if they are both 90°. Transversal angles are essentially supplementary angles on the transversal line (the line that intersects through the set of parallel lines)

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Find the area of the circle. Round your answer to the nearest hundredth.<br> 00<br> 3 in.

MAXImum [283]

Answer:

Step-by-step explanation:300

4 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

PLS HELP!!! BRAINLIEST

Alina [70]

Answer:

Using the visual, 20 blocks, but you still need to change it to match the problem's scale

Step-by-step explanation:

First just try to find a way out and count the blocks. That's what I did. The part I haven't done is scale it. As the starting and end points of the grid are given, but where they fall is not clear, I don't know what the block to number ratio is. Sorry, maybe ask a teacher, they shouldn't take off any points.

5 0

2 years ago

For the functions f(x)=2x^2+3x+9 and g(x)=−3x+10 find (f⋅g)(x) and (f⋅g)(1)

pashok25 [27]

Step-by-step explanation:

f(x)=2x²+3x+9

g(x) = - 3x + 10

In order to find (f⋅g)(1) first find (f⋅g)(x)

To find (f⋅g)(x) substitute g(x) into f(x) , that's for every x in f (x) replace it by g (x)

We have

(f⋅g)(x) = 2( - 3x + 10)² + 3(- 3x + 10) + 9

Expand

(f⋅g)(x) = 2( 9x² - 60x + 100) - 9x + 30 + 9

= 18x² - 120x + 200 - 9x + 30 + 9

Group like terms

(f⋅g)(x) = 18x² - 120x - 9x + 200 + 30 + 9

(f⋅g)(x) = 18x² - 129x + 239

To find (f⋅g)(1) substitute 1 into (f⋅g)(x)

That's

(f⋅g)(1) = 18(1)² - 129(1) + 239

= 18 - 129 + 239

We have the final answer as

Hope this helps you

3 0

4 years ago

9y^-5/3y^2r^-3 what is the answer in a fraction

allsm [11]

Answer:

3r^3/y^7.

Step-by-step explanation:

9y^-5 / 3y^2r^-3

9 / 3 = 3

y ^-5 / y^2 = y^-7 = 1 / 7y

1 / r^-3 = r^3

So the answer is 3r^3/y^7.

8 0

3 years ago