The approximate value of 400,000 multiplied by 200

A circle with radius of 5 cm sits inside a 11 cm * 11 cm rectangle.

Nadusha1986 [10]

Answer:

42.43 $cm^{2}$

Step-by-step explanation:

Given:

A circle with a radius of 5 cm sits inside a 11 cm x 11 cm rectangle.

Question asked:

What is the area of the shaded region as shown in the figure ?

Solution:

First of all calculate area of circle and then area of rectangle:-

$Area\ of\ cirle=\pi r^{2}$

$=\frac{22}{7} \times5\times5\\ \\ =\frac{550}{7}\\ \\=78.571\ cm^{2}\\$

As here length an width both are 11 cm, we can say that this is a square.

$Area\ of\ square=(Side)^{2}$

$=11\times11=121\ cm^{2}$

Now, area of shaded region = Area of square - Area of circle

= 121 - 78.571 = 42.429 $cm^{2}$

Therefore, the area of the shaded region will be 42.43 $cm^{2}$

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5 0

3 years ago

Consider a car with a price of $ 15000 that is advertised at a monthly payment of $400.00 for 48 months. What is the A.P.R?

MakcuM [25]

Answer:This would be 1

Because 1 plus 1 equals 2

Step-by-step explanation:

Bubblegum fight her If she has a crush on her If she doesn't have

4 0

2 years ago

Solve the following system of equations algebraically. Verify your solution either graphically or by using matrices.

LiRa [457]

3x - y = 0 . . . (1)
5x + 2y = 22 . . . (2)

From (1), y = 3x . . . (3)

Putting (3) into (2) gives:
5x + 2(3x) = 22
5x + 6x = 22
11x = 22
x = 2

From (3), y = 3(11) = 33

x = 11, y = 33.

5 0

2 years ago

Patrick has just finished building a pen for his new dog. The pen is 3 feet wider than it is long. He also built a doghouse to p

zalisa [80]

General Idea:

(i) Assign variable for the unknown that we need to find

(ii) Sketch a diagram to help us visualize the problem

(iii) Write the mathematical equation representing the description given.

(iv) Solve the equation by substitution method. Solving means finding the values of the variables which will make both the equation TRUE

Applying the concept:

Given: x represents the length of the pen and y represents the area of the doghouse

Statement 1: "The pen is 3 feet wider than it is long"

$Length \; of\; the \; pen = x\\ Width \; of\; the\; pen=x+3$

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Statement 2: "He also built a doghouse to put in the pen which has a perimeter that is equal to the area of its base"

$Area \; of\; the\; Dog \; house=y\\ Perimeter \; of\; Dog\; house=y$

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Statement 3: "After putting the doghouse in the pen, he calculates that the dog will have 178 square feet of space to run around inside the pen."

$Area \; of \; the\; Pen - Area \;of \;the\;Dog \;House=\;Space\;inside\;Pen\\ \\ x \cdot (x+3)-y=178\\ Distributing \;x\;in\;the\;left\;side\;of\;the\;equation\\ \\ x^2+3x-y=178\Rightarrow\; 1^{st}\; Equation\\$

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Statement 4: "The perimeter of the pen is 3 times greater than the perimeter of the doghouse."

$Perimeter\; of\; the\; Pen=3\; \cdot \; Perimeter\; of\; the\; Dog\; House\\ \\ 2(x \; + \; x+3)=3 \cdot y\\ Combine\; like\; terms\; inside\; the\; parenthesis\\ \\ 2(2x+3)=3y\\ Distribute\; 2\; in\; the\; left\; side\; of\; the\; equation\\ \\ 4x+6=3y\\ Subtract \; 6\; and \; 3y\; on\; both\; sides\; of\; the\; equation\\ \\ 4x+6-3y-6=3y-3y-6\\ Combine\; like\; terms\\ \\ 4x-3y=-6 \Rightarrow \; \; 2^{nd}\; Equation\\$