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jok3333 [9.3K]
3 years ago
5

what is the length of a hypotenuse triangle with two 45 degree angles, one 90 degree angle, and two side lengths of 3 times the

square root of 2?​
Mathematics
1 answer:
nordsb [41]3 years ago
4 0

Answer:

4.24

Step-by-step explanation:

remember a² + b² = c²

(3*(sqrt(2))² + (3*(sqrt(2))² = 18

this would be c²

square root of this is c = 4.2426...

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Step-by-step explanation:

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Find the Surface Area of the following figure. Round your answer to the nearest whole number.
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5 0
3 years ago
Need answer quickly! thank you in advance!
anyanavicka [17]

Answer:

b)(b²-a²)

Step-by-step explanation:

a cotθ + b cosecθ =p

b cotθ + a cosecθ =q

Now,

p²- q²

=(a cotθ + b cosecθ)² - (b cotθ + a cosecθ)²     [a²-b²=(a+b)(a-b)]

=(acotθ+bcosecθ + bcotθ+ acosecθ) (a cotθ + bcosecθ -bcotθ-acosecθ)

={a(cotθ+cosecθ)+b(cotθ+cosecθ)} {a (cotθ-cosecθ)+b (cosecθ-cotθ)}

={a(cotθ+cosecθ)+b(cotθ+cosecθ)} [a (cotθ-cosecθ) + {- b (cotθ-cosecθ)} ]

={a(cotθ+cosecθ)+b(cotθ+cosecθ)} {a (cotθ-cosecθ) - b (cotθ-cosecθ)}

={(cotθ+cosecθ)(a+b)} {(cotθ-cosecθ) (a-b)}

=(cotθ+cosecθ) (a+b) (cotθ-cosecθ) (a-b)

=(cotθ+cosecθ) (cotθ-cosecθ) (a+b) (a-b)        

= (cot²θ-cosec²θ) (a²-b²)                                 [(a+b) (a-b)= (a²-b²)]

= -1 . (a²-b²)                               [ 1+cot²θ=cosec²θ ; ∴cot²θ-cosec²θ=-1]

=(b²-a²)

6 0
4 years ago
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