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sleet_krkn [62]
4 years ago
9

Someone help please ;-; you actually have to prove it, not just say it bisects into 2 pieces -.-

Mathematics
2 answers:
erma4kov [3.2K]4 years ago
8 0

Answer:

hence..... proved.

be brainly........ hope it's helpful 4 u

OverLord2011 [107]4 years ago
4 0

Answer:

See below    ↓

Step-by-step explanation:

<u>Given:</u>

∠N = ∠K = 90°

LM bisects ∠KLN such that ∠KLM ≅ ∠NLM

<u>To prove:</u>

ΔKLM ≅ ΔNLM

<u>Proof:</u>

<u>Statements</u>                      |            <u>Reasons</u>

ΔKLM ↔ ΔNLM                |

∠N ≅ ∠K                           | Given (Both equal to 90°)

∠KLM ≅ ∠NLM                 | Given (Bisected Angles)

LM ≅ LM                           |   Common

ΔKLM ≅ ΔNLM                 |    A.A.S  Postulate

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If you can prove this i will mark you BRAINLIEST <br><br>sin²Acot²A+cos²Atan²A=1​
lubasha [3.4K]

Answer:

Step-by-step explanation:

cot = cos / sin

tan = sin / cos

=> sin²A × cos²A/sin²A + cos²A × sin²A/cos²

=> cos²A + sin²A

=> 1

3 0
3 years ago
If (x+2) is a factor of 3x^2+14x+k, what is the value of k?
valkas [14]
If (x+2) is a factor of 3x^2+14x+k, the equation must equal zero when x=-2 so:

3(-2)^2+14(-2)+k=0

12-28+k=0

-16+k=0

k=16
3 0
4 years ago
What is the value of the product (3-2i)(3+2i)?
Helga [31]

Answer:

13

Step-by-step explanation:

(3-2i)(3+2i) = 9  + 6i - 6i - 4i^2

9 - 4i^2

9 + 4 = 13

5 0
4 years ago
Read 2 more answers
ANSWER ASAP PLS..........
Korolek [52]

The length of one leg of the right triangle is 5√10

<h3>How to find the leg of a right triangle?</h3>

The leg of a right triangle can be found as follows;

Therefore, using trigonometric ratios,

sin 45 = opposite / hypotenuse

sin 45 = x / 10√5

1 / √2 = x / 10√5

cross multiply

10√5 = x√2

divide both sides by √2

x = 10√5 / √2

x = 10√5 / √2 × √2 / √2

x = 10√10 / 2 = 5√10

learn more on right triangle here: brainly.com/question/3398476

#SPJ1

8 0
2 years ago
What's the answer people. Help
Citrus2011 [14]
This is true <span>in the sense of cardinality. In the same sense that one can say that there are more real numbers than there are rationals (or integers; or natural numbers).</span><span>

source: </span>https://math.stackexchange.com/questions/10333/are-there-many-more-irrational-numbers-than-rational
7 0
3 years ago
Read 2 more answers
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