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ELEN [110]
4 years ago
8

Triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C'.

Mathematics
1 answer:
frozen [14]4 years ago
7 0

Note: Consider we need to find the vertices of the triangle A'B'C'

Given:

Triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C'.

Triangle A,B,C with vertices at A(-3, 6), B(2, 9), and C(1, 1).

To find:

The vertices of the triangle A'B'C'.

Solution:

If triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C', then

(x,y)\to (y,-x)

Using this rule, we get

A(-3,6)\to A'(6,3)

B(2,9)\to B'(9,-2)

C(1,1)\to C'(1,-1)

Therefore, the vertices of A'B'C' are A'(6,3), B'(9,-2) and C'(1,-1).

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A line with a slope of – 5 passes through the points (6,r) and (7, – 9). What is the value of r?
const2013 [10]

Answer:

r=-4

Step-by-step explanation:

gradient =  \frac{change \: in \: y}{change \: in \: x}  \\  - 5 =  \frac{r -  - 9}{6 - 7}  \\  - 5 =  \frac{r + 9}{ - 1}   \\  - 5 \times  - 1 = r + 9 \\ 5 = r + 9 \\ 5 - 9 = r \\ r =  - 4

Hope that this is helpful.

8 0
3 years ago
I WILL GIVE BRAINLIEST FOR THE CORRECT ANSWER AND POINTS!
iragen [17]

Answer:

36.5

Step-by-step explanation:

s² = [∑(min value - median value)² + (max value - median value)²] / (N - 1)

s² = [(23 - 59.5)² + (96 - 59.5)²] / (3 - 1)

s² = 2664.5 / 2

s² = 1332.25

s = √1332.25

s = 36.5

7 0
3 years ago
From the airport hanger H, an areoplane A is 18km away on the bearing of 115° while another aeroplane B is 29km away on a bearin
maks197457 [2]

The difference in distance between Aero plane A and B is; 34.67 km

<h3>How to calculate bearing?</h3>

To get the bearing;

∠H = (115 - 90) + (270 - 203)

∠H = 92°

Then, we will use cosine rule to get the distance between both Planes A and B.

d_ab = √(18² + 29² - 2(18 * 29) * cos 92)

d_ab = √(324 + 841 + 36.435)

d_ab = 34.67 km

Read more about bearing at; brainly.com/question/22518031

#SPJ1

6 0
2 years ago
Determine which of the following terms are not considered to be like terms with the expression -6s ( the two is squared like on
Iteru [2.4K]

Answer:

s^2t^2

4(s\cdot t)

-6(s^2+t)

Step-by-step explanation:

We want to select all the terms that are not considered to be like terms with -6s^2t.

The terms that are like terms with -6s^2t must have s^2t.

It doesn't matter the coefficient.

So we can easily see that all the following are not like terms with -6s^2t:

s^2t^2

4(s\cdot t)

-6(s^2+t)

4 0
3 years ago
Which of the following point is located on the line represented by the equation y + 4 = -5 (x - 3)?. .
horsena [70]
If we plug in: x = 3, y= - 4;
- 4 + 4 = - 5 · ( 3 - 3 )
0 = 0 ( correct )
Answer: C ) ( 3, - 4 )

3 0
3 years ago
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