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Valentin [98]
3 years ago
10

What set of numbers can represent the side lengths,in centimeters, of a right triangle

Mathematics
1 answer:
il63 [147K]3 years ago
5 0
We now look at a right-angled triangle with sides a, b and c, as shown opposite. Pythagoras' Theorem states that, for any right-angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the two shorter sides.
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Need help on this question
kifflom [539]

Answer:

C

Step-by-step explanation:

the answer for this question is

-2 and 6

5 0
3 years ago
When adding 51.2 to a certain number, the sum is 56.05, as seen below. What number should go in the box to complete the addition
Rom4ik [11]
4.85 is the answer because if you subtract 51.2 from 56.05 you get your answer
3 1
3 years ago
To decrease an amount by 18%, what single multiplier would you use?
suter [353]
I believe you would calculate this like so;

We want 18% less than 100% of that amount therefore the single multiplier we would use is as follows:

(100% - 18%) = 82% 

Which alternatively = 0.82

Does that make sense? If not I can explain it a bit more. Please let me know. 
8 0
3 years ago
Read 2 more answers
Use the data from the table to create a scatter plot. plot al the data points from the table.
aleksley [76]

Answer:

Place a point at

8,4

5,7

6,5

10,4

2,11

4,7

3,9

1,13

1,11

3,12

4,2

7,2

5,24

9,3

5,5

Step-by-step explanation:

5 0
3 years ago
EASY BRAINLIEST PLEASE HELP!!
butalik [34]

Answer:

see below

Step-by-step explanation:

<h3>Proposition:</h3>

Let the diagonals AC and BD of the Parallelogram ABCD intercept at E. It is required to prove AE=CE and DE=BE

<h3>Proof:</h3>

1)The lines AD and BC are parallel and AC their transversal therefore,

\displaystyle  \angle DAC =  \angle ACB \\  \ \qquad [\text{ alternate angles theorem}]

2)The lines AB and DC are parallel and BD their transversal therefore,

\displaystyle  \angle BD C=  \angle ABD \\  \ \qquad [\text{ alternate angles theorem}]

3)now in triangle ∆AEB and ∆CED

  • \displaystyle \angle EAD=\angle ECB
  • \angle EDA=\angle EBC
  • \displaystyle AD=BC

therefore,

\displaystyle  \Delta AEB  \cong  \Delta CED

hence,

  • AE=CE
  • DE=BE

Proven

6 0
3 years ago
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