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Ksivusya [100]
2 years ago
11

Describe 2 ways to find the area of the shaded rectangle

Mathematics
1 answer:
sweet-ann [11.9K]2 years ago
3 0
Calculate the area of both shapes. The area of a rectangle is determined by multiplying its length times its width. The area of a circle is Pi (i.e., 3.14) times the square of the radius. Find the area of the shaded region by subtracting the area of the small shape from the area of the larger shape.
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the weight of and elephant is 10,000 time the weight of a cat. If the elephant weighs 14,000 pounds, how many pounds does the ca
schepotkina [342]

Easy, the cat weighs 1.4 pounds :)

3 0
3 years ago
22. Determine the value of c so that the line
tiny-mole [99]

Answer:

If the lines BC and DE are parallel, the value of c is c=-2

Step-by-step explanation:

We are given line segment BC with end points B(2, 2) and C(9,6) and line segment DE with endpoints  D(c, -7) and E(5, -3).

Using slope formula: Slope=\frac{y_2-y_1}{x_2-x_1} we can find point c

When 2 lines are parallel there slope is same.

So, Slope of line BC =Slope of Line DE

\frac{y_2-y_1}{x_2-x_1}=\frac{y_2-y_1}{x_2-x_1}

We have:

x_1=2, y_1=2, x_2=9, y_2=6 \ for \ line \ BC \ and \\\x_1=c, y_1=-7, x_2=5, y_2=-3 \ for \ line \ DE \

Putting values and finding c

\frac{6-2}{9-2}=\frac{-3-(-7)}{5-c}\\ \frac{4}{7}=\frac{-3+7}{5-c} \\ \frac{4}{7}=\frac{4}{5-c} \\Cross \ multiply:\\4(5-c)=4*7\\20-4c=28\\-4c=28-20\\-4c=8\\c=\frac{8}{-4}\\c=-2

So, If the lines BC and DE are parallel, the value of c is c=-2

8 0
3 years ago
What is 4xy - 5y² - 3x² from 5x + 3y² - xy ? ​
andre [41]

Refer the attachment for your answer

6 0
2 years ago
Read 2 more answers
This the question complete the table below to show how to find the variance. After finding the variance, calculate the standard
nordsb [41]

Answer:

Step-by-step explanation:ik

5 0
3 years ago
HELLP!!
Naddik [55]

Answer:

(x - 2)²/2² + (y + 1)²/3² = 1 ⇒ The bold values and signs are the answers

Step-by-step explanation:

* Lets revise the equation of the ellipse  

- The standard form of the equation of an ellipse with center (h , k)  

 and major axis parallel to x-axis is (x - h)²/a² + (y - k)²/b² = 1  

- The coordinates of the vertices are (h ± a , k)

- To change the form of the equation of the ellipse to standard form we

 will using the completing square

∵ The equation of the ellipse is 9x² + 4y² - 36x + 8y + 4 = 0

- Lets collect x in bracket and y in bracket

∴ (9x² - 36x) + (4y² + 8y) + 4 = 0

- We will take a common factor 9 from the bracket of x and 4 from the

 bracket of y

∴ 9(x² - 4x) + 4(y² + 2y) + 4 = 0

- Lets make 9(x² - 4x) a completing square

∵ √x² = x ⇒ the 1st term in the bracket

∵ 4x ÷ 2 = 2x ⇒ the product of the 1st and 2nd terms

∵ 2x ÷ x = 2 ⇒ the 2nd term in the bracket

∴ The bracket is (x - 2)²

∵ (x - 2)² = x² - 4x + 4 ⇒ we will add 4 in the bracket and subtract 4

  out the bracket

∴ 9[(x² - 4x + 4) - 4] = 9[(x - 2)² - 4]

- Lets make 4(y² + 2y) a completing square

∵ √y² = y ⇒ the 1st term in the bracket

∵ 2y ÷ 2 = y ⇒ the product of the 1st and 2nd terms

∵ y ÷ y = 1 ⇒ the 2nd term in the bracket

∴ The bracket is (y + 1)²

∵ (y + 1)² = y² - 2y + 1 ⇒ we will add 1 in the bracket and subtract 1

  out the bracket

∴ 4[(y² + 2y + 1) - 1] = 4[(y + 1)² - 1]

- Lets write the equation with the completing square

∴ 9[(x - 2)² - 4] + 4[(y + 1)² - 1] + 4 = 0 ⇒ simplify

∴ 9(x -2)² - 36 + 4(y + 1)² - 4 + 4 = 0 ⇒ add the numerical terms

∴ 9(x - 2)² + 4(y + 1)² - 36 = 0 ⇒ add 36 to both sides

∴ 9(x - 2)² + 4(y + 1)² = 36 ⇒ divide both sides by 36

∴ (x - 2)²/4 + (y + 1)²/9 = 1

∵ 4 = 2² and 9 = 3²

∴ (x - 2)²/2² + (y + 1)²/3² = 1

* The standard form of the equation of the ellipse is

  (x - 2)²/2² + (y + 1)²/3² = 1

6 0
3 years ago
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