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3 years ago

What is the slope of the line on the graph? Enter your answer in the box.

Mathematics

1 answer:

Dafna1 [17]3 years ago

5 0

Answer:

-1/3x+4

Hope This Helps!!!

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Given a and b are first quadrant angles, sin a=5/13 and cos b=3/5 evaluate sin (a-b)1) -33/652) 33/653) 63/65

bezimeni [28]

The correct option is -33/65.

The solution is given below:

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1 year ago

Order from least to greatest 0.9,7/8,0.86

LenKa [72]

.86, 7/8, .9 is least to greatest

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4 years ago

What is the sum of the interior angles of a convex nonagon?

Artemon [7]

Answer:

1,260°

Step-by-step explanation:

Nonagon is a 9 sided polygon

Exterior angle = 360 ÷ 9

= 40

Interior angle + exterior angle = 180

Interior angle = 180 - 40

= 140°

Sum of interior angles = 140° × 9

= 1,260°

6 0

4 years ago

Find the perimeter AND area of the parallelogram ABCD. Show all work

stiks02 [169]

<u>Answer-</u>

<em>The perimeter and area of the parallelogram are 19.74 units and 15 sq. units respectively.</em>

<u>Solution-</u>

The co-ordinates of the vertices are,

A = (-2, 3)

B = (4, 0)

C = (1, -1)

D = (-5, 2)

E = (-3, 1)

We can get the side length of the parallelogram by calculating the respective distances by applying distance formula,

$\overline{CD}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(-5-1)^2+(2+1)^2}=\sqrt{(-6)^2+(3)^2}=\sqrt{36+9}=\sqrt{45}=3\sqrt5$

$\overline{AD}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(-2+5)^2+(3-2)^2}=\sqrt{(3)^2+(1)^2}=\sqrt{9+1}=\sqrt{10}$

$\overline{AE}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(-2+3)^2+(3-1)^2}=\sqrt{(1)^2+(2)^2}=\sqrt{1+4}=\sqrt{5}$

Perimeter of the parallelogram ABCD is,

$=2(\overline{AD}+\overline{CD})\\\\=2(\sqrt{10}+3\sqrt5)\\\\=19.74\ units$

Area of the parallelogram ABCD is,

$=\overline{CD}\times \overline{AE}\\\\=3\sqrt5\times \sqrt{5}\\\\=3\times 5\\\\=15\ sq.unit$

7 0

3 years ago

Solve for x.<br> 5x – 3 = 2x +12

mario62 [17]

Answer:

X = 5

<h3>HOPE IT IS HELPFULLY.</h3>

7 0

3 years ago