Ask question

Ask question

All categories

1 year ago

8

The CEO of NJAD, a Real Estate Investment Fund, decided to design a new development inPolycity, New Jersey. She wants to build 3

models as followsModel A has a swimming poolModel C is a parallelogram with parallel and congruent opposite sidesModel B is a trapezoid with a large backyarda) Find the perimeter of Model A above.

1 answer:

shutvik [7]1 year ago

5 0

The perimeter (P) is the sum of all sides:

$\begin{gathered} (2x^2-6x+25)+(5x^3+6x^2-3x+4)+(x^3-2x^2+8x+9)+(3x^4+2x^3+8x+7) \\ 2x^2-6x+25+5x^3+6x^2-3x+4+x^3-2x^2+8x+9+3x^4+2x^3+8x+7 \\ 3x^4+5x^3+x^3+2x^3+2x^2+6x^2-2x^2-6x-3x+8x+8x+25+4+9+7 \\ 3x^4+8x^3+6x^2+7x+45 \end{gathered}$

You might be interested in

Birth weights are normally distributed with a mean of 3419 g and a standard deviation of 494 g. If a hospital plans to set up sp

Bas_tet [7]

Let X represent a random variable representing the birth weights of babies, and let x be the required cut-off, then
P(X < x) = 0.02
P(z < (x - 3419)/494) = 0.02
1 - P(z < (3419 - x)/494) = 0.02
P(z < (3419 - x)/494) = 1 - 0.02 = 0.98
(3419 - x)/469 = 2.04
3419 - x = 956.76
x = 3419 - 956.76 = 2462.24

Therefore, the weight to be used as cut-off is 2462.24 g

3 0

3 years ago

Alex went to the grocery store and bought 5 avocados. He paid $10 and received $4.50 in change. How much did each avocado cost i

Delvig [45]

Answer:

$1.10

Step-by-step explanation:

We have been that Alex went to the grocery store and bought 5 avocados. He paid $10 and received $4.50 in change.

First of all, we will find amount paid for 5 avocados by subtracting $4.50 from $10.

$\text{Amount paid for 5 avocados}=\$10-\$4.50$

$\text{Amount paid for 5 avocados}=\$5.50$

Now, we will divide $5.50 by 5 to find cost of each avocado.

$\text{Cost of each avocado}=\frac{\$5.50}{5}$

$\text{Cost of each avocado}=\$1.10$

Therefore, the cost of each avocado is $1.10.

3 0

3 years ago

The volume of a cube can be found using the equation V = s^3, where V is the volume and s is the measure of one side of the cube

hoa [83]

Volume = Length x Length x Length

Volume = Length³

Length = ∛Volume

Given that Volume = 756 in³

Length = ∛756

Answer: ∛756 (option A)

3 0

3 years ago

A dolphin jumped up out of the water and back into the water in a parabolic path. (H) height (t) seconds . H=-8(t-0.5)^2+2 . How

tigry1 [53]

Answer:

4 seconds

Step-by-step explanation:

using the vertex formula of a quadratic,

$a(x-h)^{2} + k$ , where (h,k) is the vertex

$h = -8(t-0.5)^{2}+2$ h is height and t is time in seconds

the vertex (maximum height) of the dolphin is (h,k) or (0.5, 2)

Height of 1/2

time of 2 seconds

it will take 2 additional seconds to reach the water again.

this can also be solved using quadratic equation, but since it was already set up in vertex form, i'd use that.

4 0

3 years ago

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows

4 0

4 years ago