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3 years ago

6

Find the area of this shape

1 answer:

Dovator [93]3 years ago

5 0

I think the correct answer is 64

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Is this a function? -3x +8=y

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Yes it passes the vertical line test

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3 years ago

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Determine the graph of the polar equation 9/1-3sin theta

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Answer: can you show us the graph?

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2 years ago

Write the equation of a line for the graph:

zzz [600]

Answer: y=1/3x-2

Step-by-step explanation:

To find the equation of the line, we need the slope and y-intercept. To find the slope, we can take any two points on the graph and use the formula $m=\frac{y_2-y_1}{x_2-x_1}$ . To find the y-intercept, we find the point that crosses the y-axis. The two points on the graph are (0,-2) and (3,-1).

$m=\frac{-1-(-2)}{3-0} =\frac{1}{3}$

Now, we know the slope is 1/3.

The graph crosses the y-axis at (0,-2). Therefore the y-intercept is -2. This tells us that the equation is y=1/3x-2.

6 0

3 years ago

Problems with Equations:

Bas_tet [7]

Answer:

x = 11

The value "11" makes the equation true.

2 ( x - 5 ) = 12

2 ( 11 - 5 ) = 12

2 ( 6 ) = 12

12 = 12

Step-by-step explanation:

5 0

3 years ago

Quadrilateral ABCD has vertices A (-3, 0), B (2, 4), C (3, 1), and D (-4, -3). Calculate the perimeter of the quadrilateral. Rou

Andru [333]

Answer:

Perimeter of the Quadrilateral=20.78

Step-by-step explanation:

Perimeter of the Quadrilateral= AB+BC+CD+AD

Finding the sides of the quadrilateral using the Distance formula:

A(-3,0), B(2,4), C(3,1) and D(-4,-3)

$AB=\sqrt{(2-(-3))^2+(4-0)^2}\\\\ =\sqrt{(2+3)^2+4^2}\\\\ =\sqrt{5^2+4^2} \\\\=\sqrt{25+16}\\\\ =\sqrt{41}\\$

$=6.40$

BC=

$\sqrt{(3-2)^2+(1-4)^2} \\\\=\sqrt{1^2+(-3)^2} \\\\=\sqrt{1+9}\\\\ =\sqrt{10}$

$=3.16$

$CD=\sqrt{(-4-3)^2+(-3-1)^2} \\\\=\sqrt{(-7)^2+(-4)^2} \\\\=\sqrt{49+16}\\\\=\sqrt{65}\\\\ =8.06$

$AD=\sqrt{(-4-(-3))^2+(-3-0)^2} \\\\=\sqrt{(-4+3)^2+(-3)^2} \\\\=\sqrt{(-1)^2+(-3)^2} \\\\=\sqrt{1+9}\\\\=\sqrt{10} \\\\ =3.16$

Perimeter of the quadrilateral= AB+BC+CD+AD

=6.40+3.16+8.06+3.16

=20.78

6 0

3 years ago