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2 years ago

1. Which graph shows y as a function of x?

Mathematics

1 answer:

katen-ka-za [31]2 years ago

8 0

Answer:

Step-by-step explanation:

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Area of a trapizium is 84m².its pararrell sides are 12m and 9m. find its hiegh

Maksim231197 [3]

Answer:

8m

Step-by-step explanation:

We can convert the trapezium into a rectangle to make finding the area easier.

Say that the top side is 9m, and that the bottom side is 12m.

12 - 9 = 3

3/2* = 1.5

*Note that we are dividing by two to get the "overhang" of the bottom side.

Now that we have our overhang, we can find the length of the "t-rect"**.

**"T-rect" refers to a rectangle that has been formed by changing a trapezium into a rectangle.

12 - 1.5 = 10.5

9 + 1.5 = 10.5

The length of the t-rect is 10.5.

84/10.5 = 8

The height of the t-rect is 8m.

Since the height of the original trapezium is equal to the height of the t-rect***,

***Since we have only adjusted the length

this also means that 8m would also be the height of the original trapezium.

Now, there is a second way of doing this.

Find the median of 12 and 9.

9, 10, 11, 12

10, 11

The median is 10.5.

84/10.5 = 8

You get the same answer. These methods could be used as a solve-then-check method to another problem like these.

Hope this helps. Have a nice day.

3 0

3 years ago

PLEASE HELP EMERGENCY !! PLS PLS

tatyana61 [14]

Call 911 then but the answer is $4 because 12/3 is 4

4 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

Marsha wants to buy carpet to cover her whole living room, except for the tiled floor. The tiled

White raven [17]

Answer:

68.06 sq. feets

Step-by-step explanation:

Area of room's floor

11.33 × 7 = 79.33

Area of tiled floor

4.83 × 2.33 = 11.278

Area of floor needed to be carpeted

79.33-11.278= 68.06

6 0

3 years ago

Evaluate the integral Integral ∫ from (1,2,3 ) to (5, 7,-2 ) y dx + x dy + 4 dz by finding parametric equations for the line seg

n200080 [17]

$\vec F(x,y,z)=y\,\vec\imath+x\,\vec\jmath+3\,\vec k$

is conservative if there is a scalar function $f(x,y,z)$ such that $\nabla f=\vec F$ . This would require

$\dfrac{\partial f}{\partial x}=y$

$\dfrac{\partial f}{\partial y}=x$

$\dfrac{\partial f}{\partial z}=3$

(or perhaps the last partial derivative should be 4 to match up with the integral?)

From these equations we find

$f(x,y,z)=xy+g(y,z)$

$\dfrac{\partial f}{\partial y}=x=x+\dfrac{\partial g}{\partial y}\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)$

$f(x,y,z)=xy+h(z)$

$\dfrac{\partial f}{\partial z}=3=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=3z+C$

$f(x,y,z)=xy+3z+C$

so $\vec F$ is indeed conservative, and the gradient theorem (a.k.a. fundamental theorem of calculus for line integrals) applies. The value of the line integral depends only the endpoints:

$\displaystyle\int_{(1,2,3)}^{(5,7,-2)}y\,\mathrm dx+x\,\mathrm dy+3\,\mathrm dz=\int_{(1,2,3)}^{(5,7,-2)}\nabla f(x,y,z)\cdot\mathrm d\vec r$

$=f(5,7,-2)-f(1,2,3)=\boxed{18}$

8 0

3 years ago

1. Which graph shows y as a function of x?​

1. Which graph shows y as a function of x?