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Sedaia [141]
1 year ago
7

HiCan you please state the Pythagorean TheoremVerbally and algebraically

Mathematics
1 answer:
Natalija [7]1 year ago
7 0
Pythagorean Theorem<h2>Verbally:</h2>

Let's say a and b are the legs, and c is the hypotenuse. Then, algebraically, the theorem is,

a^2+b^2=c^2

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5/6 b + 5 - 2/3 b -2
Oksana_A [137]

Answer:

1/6 b+3

Step-by-step explanation:

6 0
3 years ago
What are the solutions to the expression x^2-8x=24
labwork [276]

Answer: x= 2(2+sqrt(10)) , x=2(2-sqrt(10)

Step-by-step explanation:

x^2-8x=24 <- subtract 24 from both sides

x^2-8x-24 <- take quadratic formulat (-2+/- Sqrt(b^2 - 4ac)/2a

-(-8) +/- sqrt((-8)^2 - 4(1*-24) all over (2*1) <- simplify

(-(-8) +/- 4sqrt(10) )/2 <- sepereate

(-(-8) +/4sqrt(10) )/2 ,  (-(-8) - 4sqrt(10) )/2 <- simplify

2(2+sqrt(10)) , 2(2-sqrt(10) =x

6 0
3 years ago
How to solve and what is the answer
TiliK225 [7]
Two ways: 
1) guess factors(trial and error)
2) use quadratic formula.
If you use this method then a = -3, b = -6 and c = -1


x = -b +/- [sqrt(b^2 -4ac)/2a]
substituting a, b, and c into our equation we get:
x = - (-6)+/- [sqrt ((-6)^2) - 4(-3)(-1))/2 (-3)]
x = + 6 +/- [sqrt (36 -4 (3)/-6)]  if I didnt make a mistake in my signs
x = + 6 +/- [sqrt (36 -12)/-6)]
x = 6 +/- [sqrt (24)/-6]  but sqrt 6 x sqrt of 4 = sqrt 24 hence
x = 6 +/-  [ sqrt 6 x sqrt 4 /-6] that is:
x = 6 +/- [sqrt 6 x 2 /-6 ]
so x = 6 + [sqrt 6 x 2/-6] and x = 6 - [sqrt 6 x 2/-6]

7 0
3 years ago
Find the image of the given point under the given translation!!!! help please!!!!​
harkovskaia [24]

Answer:

P' = (7, - 8)

Step-by-step explanation:

Under the given translation (x + 4, y - 3)

Add 4 to the x- coordinate of P and subtract 3 from the y- coordinate of P

P' = (3 + 4, - 5 - 3 ) = (7, - 8)

7 0
3 years ago
Decrease £900 by 5%?
GaryK [48]
45 should be the answer for that specific question.
4 0
3 years ago
Read 2 more answers
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