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xeze [42]
3 years ago
10

(Notimated as brainliest 25 points )

Mathematics
1 answer:
Genrish500 [490]3 years ago
7 0

Answer:

Triangle ABC is not a right triangle because d) it doesn't have a pair of perpendicular sides

Step-by-step explanation:

Perpendicular sides mean that the slopes are "opposite reciprocals" of each other.

Example: slope is 1/2 and the perpendicular line's slope is -2

How? 1/2 changes to 2 (you flip it, reciprocal), and then from 2 to -2 (apply opposite sign)

So when you calculate the slopes of AB, AC, and BC, none of their slopes are opposite reciprocals of each other. You can calculate the slopes of each and compare them if you want to check.

I really hope this helps, do you think you can tell me if you got the answer correct? I'd feel really bad if you got it wrong because of me. All I know is that it's not a right triangle.

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Plz help i just dont know plz help
Tpy6a [65]

Answer:

1. 266\dfrac{2}{3}\%

2. 400\%

Step-by-step explanation:

1. Fraction \frac{1}{4} is 100%, then

\dfrac{1}{4} - 100\%\\ \\\dfrac{2}{3} - x\%

Write a proportion:

\dfrac{\frac{1}{4}}{\frac{2}{3}}=\dfrac{100}{x}

Cross multiply:

\dfrac{1}{4}x=\dfrac{2}{3}\cdot 100\\ \\x=4\cdot \dfrac{2}{3}\cdot 100\\ \\x=\dfrac{800}{3}\\ \\x=266\dfrac{2}{3}\%

2. Fraction \frac{1}{6} is 100%, then

\dfrac{1}{6} - 100\%\\ \\\dfrac{2}{3} - x\%

Write a proportion:

\dfrac{\frac{1}{6}}{\frac{2}{3}}=\dfrac{100}{x}

Cross multiply:

\dfrac{1}{6}x=\dfrac{2}{3}\cdot 100\\ \\x=6\cdot \dfrac{2}{3}\cdot 100\\ \\x=\dfrac{1,200}{3}\\ \\x=400\%

3 0
3 years ago
Multiply by _<br> 3 9<br> 10 _<br> 8 24
Viktor [21]

Answer:

Multiply by 3

10 × 3 = 30

Step-by-step explanation:

Multiply by x

3x = 9

10x = y

8x = 24

--------------------------------------

3x = 9

x = 3 --- divide both sides by 3

8x = 24

x = 3 --- divide both sides by 3

10(3)

30

6 0
3 years ago
JKL has vertices J(3, -5) K(-8, -1) and L(5,1). JKL will be reflected across the line y= -1 to form J'K'L'. Which quadrant will
katen-ka-za [31]

Answer:

The correct answer is second quadrant.

Step-by-step explanation:

Vertices of J ≡ (3 , -5) which shows J is in fourth quadrant. When J is reflected across the line y = -1, the point J moves to first quadrant as J' with coordinates (3 , 3).

Vertices of K ≡ (-8 , -1) which shows K is in third quadrant. When K is reflected across the line y = -1, the point K remains where it is (in the third quadrant) as it is on the line y = -1 only.

Vertices of L ≡ (5 , 1) which shows L is in first quadrant. When L is reflected across the line y = -1, the point L moves to fourth quadrant as L' with coordinates (5 , -3).

Thus each of the first, fourth and third quadrant contains a vertex except the second quadrant.

4 0
3 years ago
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