Consider the system of linear equations.

x + y = 9.0 0.50 x + 0.20 y = 3.30

Find the values of x and y

After converting the provided integral to polar coordinates, the value of integral is evaluated π/2.

When the Cartesian coordinates (x,y) are expressed in the polar coordinates (r, θ), then this form is called the polar form.

The given integral function in the problem is,

[tex]\int_0^{\sqrt{4}} \int\limits^x_{-x} dydx[/tex]

Let suppose, [tex]x=r\cos\theta[/tex] and [tex]y=r\sin\theta[/tex]. Thus,

[tex]\sin\theta=\dfrac{y}{r}\\\cos\theta=\dfrac{x}{r}[/tex]

Limits are  y=x. From the trigonometry, the value of theta in the given triangle can be given as,

[tex]\dfrac{\sin\theta}{\cos\theta}=1\\\tan\theta=1\\\theta=\tan^{-1}1\\\theta=45^o\\\theta=\dfrac{\pi}{4}[/tex]

Similarly, for y=-x the value of angle,

[tex]\theta=-\dfrac{\pi}{4}[/tex]

Thus, the limits of theta are from -π/4 to π/4. From the Pythagoras theorem,

[tex]r^2=x^2+y^2\\r^2=(r\cos\theta)^2+(r\sin\theta)^2\\r^2=r^2(1)[/tex]

Thus, the limits of r is from 0 to 1. Convert the given integral in polar form as,

[tex]\int\limits^{\pi/4}_{-\pi/4} \int_0^{1} dt ds\\\int\limits^{\pi/4}_{-\pi/4} [1-0] dt \\\int\limits^{\pi/4}_{-\pi/4} dt \\\dfrac{\pi}{4}-\left(-\dfrac{\pi}{4} \right) \\\dfrac{\pi}{2} \\[/tex]

Hence, after converting the provided integral to polar coordinates, the value of integral is evaluated π/2.

Learn more about the polar form here;

https://brainly.com/question/24943387

To convert the given Cartesian integral to polar coordinates, identify the bounds in polar terms, then rewrite the integral accordingly. After setting up the new limits for r and θ, use the relationship between Cartesian and polar coordinates to express the area element, and integrate step-by-step.

To convert the integral ∫4√0∫x-xdydx to polar coordinates and evaluate it, we first need to describe the limits of integration and the region of integration in terms of polar coordinates (r, θ). The given integral ranges over a region bounded by the parabola y = √x and the x-axis from x=0 to x=4. Converted to polar coordinates, this region is bounded by the rays θ = 0 and θ = π/2 and the circles r = 0 and r = 4cos(θ).

So the double integral can be rewritten as ∫π/20∫4cos(θ)0 rdrdθ. To evaluate this integral, we integrate r from 0 to 4cos(θ), then integrate θ from 0 to π/2:

∫π/20 (∫4cos(θ)0 r dr) dθ = ∫π/20 [1/2 r^2]|^{4cos(θ)}_0 dθ = ∫π/20 8cos^2(θ) dθ
Using the double angle formula, cos^2(θ) = (1+cos(2θ))/2, the integral becomes:

8 ∫π/20 (1+cos(2θ))/2 dθ = 4 ∫π/20 (1+cos(2θ)) dθ

This can now be integrated directly to get the final result.

Answer:false

Step-by-step explanation:

Integer contains only both positive and negative numbers digits but not decimal points.

Answer:False

Step-by-step explanation:The set of integers Z ={....-3,-2,-1,0,1,2,3....}

Answer:

a) positive

b)

[tex]^\circ F\text{ per Minute}[/tex]      

c) Interpretation of f'(20)=2  

Step-by-step explanation:

We are given the following in the question:

[tex]T=f(t)[/tex]

where T is the temperature in degrees Fahrenheit of a cold yam placed in a hot oven and t is the time in minutes since the yam was put in the oven.

a)  sign of f'(t)

f'(t) will represent the rate of change in temperature.

f'(t) will represent the change in temperature of yam when 1 minute has passed since it was kept in oven.

Since the temperature will always increase in oven, f'(t) will have a positive sign.

b) units of f'(20)

Since, f'(t) represent the rate of change in temperature. the unit will be

[tex]\dfrac{\text{degrees Fahrenheit}}{\text{Minute}}[/tex]

That is degrees Fahrenheit per minute.

c)  f'(20)=2

f'(20) will tell the change in temperature when 20 minutes have passed after the yam has been kept in oven.

Thus, the given statement means that 20 minutes after the yam was kept for  in the oven, the temperature of yam was increasing by 2 degree Fahrenheit per minute.

The sign of f'(t) is positive . The units of f'(20) are degrees Fahrenheit per minute. The statement f'(20)=2 means that at 20 minutes, the temperature of the yam is increasing at a rate of 2°F per minute.

The temperature, T, of a cold yam placed in a hot oven, as a function of time t is given by T=f(t), where t is the time in minutes since the yam was put in the oven.

The sign of f'(t) is positive. This is because, as time t increases, the temperature of the cold yam increases due to the hot environment of the oven.

The units of f'(20) are degrees Fahrenheit per minute (°F/min). This is because f'(t) represents the rate of change of temperature with respect to time.

The practical meaning of the statement f'(20)=2 is that at t=20 minutes, the temperature of the yam is increasing at a rate of 2 degrees Fahrenheit per minute.

Answer:

the correct answer is B. g(x)=3^x +1

Step-by-step explanation:

I just took the test

Hope this helps

The exponential function with a y-intercept of 2 is in the form y = 2b^x, where 'a' represents the y-intercept of the function. To find this, set x to 0 in the function, resulting in y = a(1), and thus y = a. If the function's y-intercept is 2, the value of 'a' is 2.

To determine which exponential function has a y-intercept of 2, you need to recall the standard form of an exponential function, y = abx, where a is the y-intercept of the function. For an exponential function, the y-intercept occurs when x is 0. Thus, when x = 0, the function takes the form y = ab0, and since anything to the power of 0 is 1, the function simplifies to y = a. Therefore, if a function's y-intercept is 2, it means that the value of a must be 2, resulting in the function y = 2bx.

Additionally, understanding the relationship between exponential and logarithmic functions can be helpful. To rewrite a base number b in terms of natural logarithms, you can use the fact that b = eln(b). For example, 2 = eln(2). This is valuable for solving equations involving exponential growth or decay, especially when a y* calculator button is unavailable.

Answer:

The area of the rectangle is [tex]A=5L-L^2[/tex].

Step-by-step explanation:

The perimeter of a rectangle is

[tex]P=2(L+W)[/tex]

where, L is length and W is width.

It is given that perimeter of a rectangle is 10m.

[tex]10=2(L+W)[/tex]

Divide both sides by 2.

[tex]5=L+W[/tex]

Subtract L from both sides.

[tex]5-L=W[/tex]

Area of a rectangle is

[tex]A=L\times W[/tex]

Substitute W=(5-L) in the above formula.

[tex]A=L\times (5-L)[/tex]

[tex]A=5L-L^2[/tex]

Therefore, the area of the rectangle is [tex]A=5L-L^2[/tex].

Final answer:

The area A of a rectangle with a fixed perimeter of 10 meters is expressed as a function of the length L by the formula A(L) = L(5 - L), assuming 0 ≤ L ≤ 5 meters.

Explanation:

The question is asking us to express the area A of a rectangle as a function of the length L, one of its sides, given a fixed perimeter of 10 meters. The perimeter of a rectangle is given by the formula P = 2L + 2W, where L is the length and W is the width. For a rectangle with a perimeter of 10 meters, we have:

2L + 2W = 10

W = (10 - 2L) / 2 = 5 - L

The area A of the rectangle is A = L × W = L(5 - L)

This formula A = L(5 - L) gives the area as a function of the length.A: The rectangle's area is a function of its length, expressed as A(L) = L(5 - L), valid for 0 ≤ L ≤ 5, since the minimum possible width is 0 when the length equals 5, and the maximum possible length is 5 when the width equals 0.

Answer:

ab dc ef is replaced by 12 43 56

Step-by-step explanation:

ab dc ef are replaced by the position they take when writing in an alphabetical order. a is 1, b is 2, c is 3, and so on.

So, ab dc ed is written as

12 43 56

and

56 - 43 - 12 = 1

Answer:

Stem-and-leaf plot of the test scores is shown below.

Step-by-step explanation:

The given data set is

67, 72, 86, 75, 89, 89, 87, 90, 99, 100

Stem-and-leaf: Leaf is the last term and stem is other term. If a number is 32, then 3 is stem and 2 is leaf.

Stem-and-leaf plot of the test scores is

Stem            leaf

6                  7

7                  2,5

8                  6,7,9,9

9                  0,9

10                 0

The length of the rows are similar to the heights of bars in a histogram; longer rows of the data correspond to higher frequency.