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2 years ago

Write an equation in standard form for the line that contains the given pair of points.

Mathematics

1 answer:

wel2 years ago

6 0

Answer:

x = 3

Step-by-step explanation:

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Someone help with math , thanks <3

NikAS [45]

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1 year ago

In ΔUVW, the measure of ∠W=90°, VW = 32 feet, and WU = 78 feet. Find the measure of ∠V to the nearest degree.

DanielleElmas [232]

Answer: $67.68^{\circ}$

Step-by-step explanation:

Given

$\angle W=90^{\circ}$

$VW=32\ ft$

$WU=78\ ft$

from the figure, we can write

$\Rightarrow \tan v=\dfrac{UW}{VW}\\\\\Rightarrow \tan v=\dfrac{78}{32}\\\\\Rightarrow \tan v=2.437\\\Rightarrow v=\tan^-1(2.437)\\\Rightarrow v=67.68^{\circ}$

7 0

2 years ago

-22 + (-16)<br> Hhhhhhhhhhhhhhhhhhhhhh

Ad libitum [116K]

Answer:

-22 + (-16)= -38

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Solve for the variable.<br> The quotient of a number n divided by -4.5 equals 200.6.

jeyben [28]

Answer:

n = -902.7

Step-by-step explanation:

n/-4.5 = 200.6

n = (-4.5)(200.6)

n = -902.7

8 0

1 year ago

Please help! Give the answer and how you got it

Ronch [10]

The distance between the points C(-3,1) and D(5,6) is 10.3 units.

<u>Step-by-step explanation:</u>

Let us consider CD is a line segment.

The point C is (-3,1) and D is (5,6).

The Cartesian formula for line is used to find the length between the CD.

Distance between two points = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ .

The points are (-3,1) and (5,6).

= $\sqrt{(6-(-3))^2+(6-1)^2}$ .

= $\sqrt{(6+3)^2+(6-1)^2}$

= $\sqrt{9^2+5^2}$ .

= $\sqrt{81+25}$ .

= $\sqrt{106}$ .

=10.295 ≈ 10.3.

Distance between two points 10.3 units.

7 0

2 years ago