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3 years ago

Write an equation in point-slope form of a line that has a slope of 2/3 and

Mathematics

1 answer:

Serggg [28]3 years ago

5 0

Answer:

y= 2/3x -1

Step-by-step explanation:

Ok so basically we want to put it in the form y=mx+b where m is 2/3 and passes through the point (9,5). So sub in the x and y of the point in brackets into the equation. 5= 2/3(9) + b and solve for b so b = -1. Therefore the equation is y=2/3x -1

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For ΔABC, ∠A = 2x - 2, ∠B = 2x + 2, and ∠C = 5x. If ΔABC undergoes a dilation by a scale factor of 1 2 to create ΔA'B'C' with ∠A

Zolol [24]

<h3>Answer:</h3>

A) ∠A = ∠A' = 38° and ∠B = ∠B' = 42°

<h3>Explanation:</h3>

The sum of angles in ∆ABC is 180°, so ...

... (2x -2) + (2x +2) + (5x) = 180

... 9x = 180

... x = 20

and the angles of ∆ABC are ∠A = 38°, ∠B = 42°, ∠C = 100°.

___

The sum of angles of ∆A'B'C' is 180°, so ...

... (58 -x) +(3x -18) +(120 -x) = 180

... x +160 = 180

... x = 20

and ∠A' = 38°, ∠B' = 42°, ∠C' = 100°.

_____

The values of angle measures of ∆ABC match those of ∆A'B'C', so we can conclude ...

... A) ∠A = ∠A' = 38° and ∠B = ∠B' = 42°

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3 years ago

Jeremy ran 1500 meters in 4 minutes, 45 seconds. How fast did he run in meters per second?

soldi70 [24.7K]

Answer:

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3 years ago

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vagabundo [1.1K]

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3 years ago

What would be the coordinates of triangle A'B'C' if triangle ABC was dilated by a factor of

notsponge [240]

Answer:

$A(0,0)\rightarrow A'(0,0) \\ B(2,0)\rightarrow B'(\frac{2}{3},0) \\ C(0,2)\rightarrow C'(0,\frac{2}{3})$

Step-by-step explanation:

The question is as following:

The verticies of a triangle on the coordinate plane are

A(0, 0), B(2, 0) and C(0, 2).

What would be the coordinates of triangle A'B'C' if triangle ABC was dilated by a factor of 1/3 ?

=============================================

Given: the vertices of a triangle ABC are A(0, 0), B(2, 0) and C(0, 2).

IF the triangle is dilated by a factor of k about the origin, then

(x,y) → (kx , ky)

that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.

It is given that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.

If a figure dilated by a factor of 1/3 about the origin

So, $(x,y)\rightarrow (\frac{1}{3}x,\frac{1}{3}y)$

<u>So, The coordinates of the triangle A'B'C' are:</u>

$A(0,0)\rightarrow A'(0,0) \\ B(2,0)\rightarrow B'(\frac{2}{3},0) \\ C(0,2)\rightarrow C'(0,\frac{2}{3})$

6 0

3 years ago

PLEASE HELP<br>what is the value of x? enter your answer in the box

Mashcka [7]

Answer: 3

For this question, we will use the angle bisector theorem.

Angle Bisector Theorem: In geometry, the angle bisector theorem is concerned with the relative lengths of the two segments that a triangle's side is divided into by a line that bisects the opposite angle. It equates their relative lengths to the relative lengths of the other two sides of the triangle.

Now let's place it according to a general formula.

x/2.25 = 4/3

x= 4/3 * 2.25

x= 3

There you go! now you know the answer and the way to do it! Brainliest pweasee if the answer is correct and helpful!

<h2>૮꒰ ˶• ༝ •˶꒱ა</h2><h2>./づᡕᠵ᠊ᡃ࡚ࠢ࠘ ⸝່ࠡࠣ᠊߯᠆ࠣ࠘ᡁࠣ࠘᠊᠊°.~♡︎ Sara Senpie</h2>

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2 years ago