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Vinvika [58]
3 years ago
12

consider the point A= (1,2) and the line B, given by the equation y=3x-5. write an equation in slope intercept of the line passi

ng through point A and parallel to line B. Please answer quickly!
Mathematics
1 answer:
Gre4nikov [31]3 years ago
5 0

Answer:

y = 3x - 1

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = 3x - 5 ← is in slope- intercept form

with slope m = 3

Parallel lines have equal slopes, thus

y = 3x + c ← is the partial equation of the parallel line

To find c substitute (1, 2) into the partial equation

2 = 3 + c ⇒ c = 2 - 3 = - 1

y = 3x - 1 ← equation of parallel line

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How much does 18cm are in mm
77julia77 [94]
Answer:
18 cm = 180 mm
Explanation:
18 cm • 10 cm (in one mm) = 180 mm
6 0
3 years ago
Find mZP.<br> P<br> 26<br> 2./13<br> R<br> mZP =<br> I’m confused
kondaur [170]

Answer:

45 degrees

Step-by-step explanation:

Since we have a right triangle, we use the appropriate trigonometric ratio

The angle marked P faces QR and that makes it the opposite side

PR faces the right angle and that makes it the hypotenuse

The trigonometric ratio that links the opposite and the hypotenuse is the sine

It is the ratio of the opposite to the hypotenuse

Thus;

sin P = √26/2 √13

Sin P = √2/2

P = arc sin √2/2

P = 45 degrees

8 0
3 years ago
(2,-4) is reflected across the x-axis. what are the coordinates of its image?
masha68 [24]

Answer:

(2, 4)

Step-by-step explanation:

When you reflect/flip over an axis, negate value of the opposite of it. Fox the X axis, negate the y value.

5 0
3 years ago
Read 2 more answers
URGENT! PLEASE HELP.
Harrizon [31]

The distance BC from Tower 2 to the plane will be 14,065.5 ft and the height of the plane from the ground will be 5,720.9 ft.

<h3>What is a right-angle triangle?</h3>

It is a type of triangle in which one angle is 90 degrees and it follows the Pythagoras theorem and we can use the trigonometry function.

A plane is located at C on the diagram.

There are two towers located at A and B.

The distance between the towers is 7600 feet and angles of elevation are given.

Then in the right-angle triangle ΔADC, we have

\rm \tan 16 = \dfrac{H}{7600 + X}\\\\\\H = 0.2867(7600 +X)\\\\\\H = 0.2756\ X + 2179.2649 ...1

Then in the right-angle triangle ΔBDC, we have

\rm \tan24 = \dfrac{H}{ X}\\\\\\H = 0.4452\ X ...2

From equations 1 and 2, we have

0.4452X = 0.2756 X + 2179.2649

0.1696X = 2179.2649

            X = 12849.439 ≅ 12,849.4 ft

Then the distance BC from Tower 2 to the plane will be

\rm BC = \dfrac{12849.4}{\cos 24}\\\\\\BC = 14065.5 \ ft

Then the height of the plane from the ground will be

\rm H = 12849.4 \times  \tan 24 \\\\\\H = 5720.9 \ ft

The figure is shown below.

More about the right-angle triangle link is given below.

brainly.com/question/3770177

#SPJ1

3 0
2 years ago
What is the length of BC ?
MrRa [10]
Looking at this triangle, we can see that is a form of a 45, 45, 90 triangle
This would mean that 2x-24=x-2
Solve: 2x-24=x-2
subtract x from both sides and add 24 to both sides...
x=22
6 0
3 years ago
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