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3 years ago

Plz help which one is not a chord

Mathematics

1 answer:

kaheart [24]3 years ago

7 0

Answer:

Line FN.

Step-by-step explanation:

A chord is a line whose endpoints both lie on the circle.

All the other lines' endpoints are on the circle.

In line FN, point N is on the circle but point F is not.

This makes line FN not a chord.

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What is the area of the piece of red paper after the hole for

MakcuM [25]

Answer:

$\large\boxed{\large\boxed{47unit^2}}$

Explanation:

The complete question is:

A hole the size of a photograph is cut from a red piece of paper to use in a picture frame.

On a coordinate plane, 2 squares are shown. The photograph has points (-3, -2), (- 2, 2), (2, 1), and (1, -3). The red paper has points (- 4, 4), (4, 4), (4, -4), and (-4, -4).

What is the area of the piece of red paper after the hole for the photograph has been cut?

<h2>Solution</h2>

The area of the piece of redpaper after the hole has been cut is equal to the area of the big square less the area of the small rectangle.

1. Area of the big rectangle

Vertices:

(- 4, 4)
(4, 4)
(4, -4)
(-4, -4)

The side length is the distance between two consecutive vertices:

The two points (-4,4) and (4,4) has the same y-coordinate, thus the length is just the absolute value of the difference of the x-coordinates:

$4-(-4)=4+4=8$

The area of this square is $(8unit)^2=64unit^2$

2. Area of the small square

Vertices:

(-3, -2)
(- 2, 2)
(2, 1)
(1, -3)

To find the side length use the distance formula for two consecutive points, for instance (2,1) and (-2,2):

$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\ \\d=\sqrt{(2-(-2))^2+(2-1)^2}=\sqrt{4^2+1^2}=\sqrt{17}$

The area is:

$(\sqrt{17}unit)^2}=17unit^2$

3. Area of the piece of red paper after the holw for the photograph has been cut:

Find the difference:

$64unit^2-17unit^2=47unit^2$

7 0

3 years ago

Read 2 more answers

Please explain to me

kolbaska11 [484]

$\bf \stackrel{\textit{product of a }\stackrel{x}{number}\textit{ and 8}}{8\cdot x}~~=~~\stackrel{\textit{104 more than 4 times that }\stackrel{x}{number}}{4\cdot x+104}\\\\\\~~~~~~~~~~~~~~~~~~~~~~8x=4x+104\\\\-------------------------------\\\\\stackrel{\textit{quotient of a }\stackrel{x}{number}\textit{ and 6}}{\cfrac{x}{6}}~~=~~\stackrel{\textit{15 less than a }\stackrel{x}{number}}{x-15}~~\implies \cfrac{x}{6}=x-15$

5 0

3 years ago

Read 2 more answers

On the map the scale is 1 inch=125 miles. What is the actual distance between the two cities if the map distance is 2 3/4 inches

ss7ja [257]

291.6 or rounded 292 miles

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3 years ago

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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

2 years ago

What property is being illustrated below? 2(7+3)=2(7)+2(3)

AleksAgata [21]

Answer: Associative property

Step-by-step explanation:

8 0

2 years ago