All categories

9 days ago

nine days ago the area covered by a mold on a piece of bread was 3 square inches today the mold covers 9,square inches find the

rate of change

Mathematics

You might be interested in

A rectangular schoolyard is to be fenced in using the wall of the school for one side and 150

lawyer [12517]

Answer:

Part 1) The equation is $A(x)=150x-2x^2$

Part 2) When x=40 m, the area of the schoolyard is A=2,800 m^2

Part 3) The valid domain consists of all real numbers exceeding zero and below 75 meters

Step-by-step explanation:

Part 1) Formulate an expression for A(x)

Let

x -----> the length of the rectangular school yard

y ---> the width of the rectangular school yard

It is known that

The perimeter for the fencing (taking the school wall as one side) is

$P=2x+y$

$P=150\ m$

thus

$150=2x+y$

$y=150-2x$ -----> this is equation A

The area of the rectangular school yard is

$A=xy$ ----> this is equation B

Substituting equation A into equation B yields

$A=x(150-2x)$

$A=150x-2x^2$

Change to function notation

$A(x)=150x-2x^2$

Part 2) What is the area when x=40?

With x equal to 40 m

substitute the value into the expression from Part 1 to determine A

$A(40)=150(40)-2(40^2)$

$A(40)=2,800\ m^2$

Part 3) What would be a suitable domain for A(x) in this scenario?

We understand that

A signifies the area of the rectangular school yard

x characterizes the length of the rectangular school yard

It follows that

$A(x)=150x-2x^2$

This forms a vertical parabola opening downwards

The vertex indicates a maximum point

The x-coordinate of the vertex corresponds to the length that maximizes the area

The y-coordinate of the vertex denotes the maximum area

The vertex corresponds to (37.5, 2812.5)

Refer to the accompanying figure

Consequently,

The peak area achieved is 2,812.5 m^2

The x-intercepts are located at x=0 m and x=75 m

The domain for A is the range -----> (0, 75)

All real numbers greater than zero and less than 75 meters

5 0

3 months ago

Gene starts from home and travels 3 miles north to the shopping mall. From the shopping mall, he travels 2 miles west to the lib

lawyer [12517]

Answer: The most significant angle created during his journey appears at the mall, between his house and the library.

Step-by-step explanation:

Hi, since this scenario forms a right triangle (refer to the attached image), the angle between his house and the library measures 90°.

For a right triangle, the total of its internal angles equals 180°, making the right angle (90°) the largest among them.

Thus, the angle at the mall, between his house and the library, is indeed the largest angle formed during his trip.

If you need further clarification or have questions, feel free to ask!

7 0

2 months ago

Read 2 more answers

The ground-state wave function for a particle confined to a one-dimensional box of length L is Ψ=(2/L)^1/2 Sin(πx/L). Suppose th

AnnZ [12381]

Respuesta:

(a) 4.98x10⁻⁵

(b) 7.89x10⁻⁶

(d) 0.5

(e) 2.9x10⁻²

Explicación paso a paso:

La probabilidad (P) de encontrar la partícula está dada por:

$P=\int_{x_{1}}^{x_{2}}(\Psi\cdot \Psi) dx = \int_{x_{1}}^{x_{2}} ((2/L)^{1/2} Sin(\pi x/L))^{2}dx$

$P = \int_{x_{1}}^{x_{2}} (2/L) Sin^{2}(\pi x/L)dx$ (1)

La solución de la integral de la ecuación (1) es:

$P=\frac{2}{L} [\frac{X}{2} - \frac{Sin(2\pi x/L)}{4\pi /L}]|_{x_{1}}^{x_{2}}$

(a) La probabilidad de encontrar la partícula entre x = 4.95 nm y 5.05 nm es:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{4.95}^{5.05} = 4.98 \cdot 10^{-5}$

(b) La probabilidad de encontrar la partícula entre x = 1.95 nm y 2.05 nm es:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{1.95}^{2.05} = 7.89 \cdot 10^{-6}$

(c) La probabilidad de encontrar la partícula entre x = 9.90 nm y 10.00 nm es:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{9.90}^{10.00} = 1.89 \cdot 10^{-4}$

(d) La probabilidad de encontrar la partícula en la mitad derecha de la caja, es decir, entre x = 0 nm y 50 nm es:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{50.00} = 0.5$

(e) La probabilidad de encontrar la partícula en el tercio central de la caja, es decir, entre x = 0 nm y 100/6 nm es:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{16.7} = 2.9 \cdot 10^{-2}$

Espero que te ayude.

3 0

2 months ago

In year 3 it is expected that the total value of clothing sales will reach 32 million if the total value of ASCO sells Remains t

tester [12383]

Answer:

The sales figure for the third year is $[\frac{32\ \text{mn}-x\ \text{mn}}{x\ \text{mn}}\times 100\%]$ .

Step-by-step explanation:

Since sales data for the second year is unavailable, we denote it as x million.

Year 3's total sales amount to 32 million.

Calculate the sales account for year 3 as follows:

$\text{Percentage of sales for year 3}=\frac{\text{Total Sales in year 3}-\text{Total Sales in year 2}}{\text{Total Sales in year 2}}\times 100\%$

$=\frac{32\ \text{mn}-x\ \text{mn}}{x\ \text{mn}}\times 100\%$

3 0

3 months ago