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Irina-Kira [14]
3 years ago
8

in a right angled triangle if the hypotenuse is 15 cm and the ratio of the other two sides is 3:4 find the measures of the right

angled triangle
Mathematics
1 answer:
wariber [46]3 years ago
7 0

Answer:

the other sides are 9 and 12

Step-by-step explanation:

use the Pythagorean theorem.

a^{2} +b^{2} =c^{2}

you know that c^2 is 225 because 15*15=225

a^{2} +b^{2} =225

(from here on I just plugged numbers in)

3:4=9:12

9^{2} +12^{2} =225

81+144=225

225=225

Yaaaaay!

(Also can I please have Brainliest? I need it to level up!)

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ziro4ka [17]

The vertical angles are congruent (or equal) so,

32 = 2x

2x = 32

2x/2 = 32/2 ..... dividing both sides by 2

x = 16 which is the answer

3 0
3 years ago
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The volleyball team is having a carwash fundraiser. The cost of each car wash is $5. They sold some season ticket packages, whic
noname [10]

Answer:

They need to wash 72 cars

Step-by-step explanation:

2000 - 1640 = 360

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5 0
2 years ago
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nasty-shy [4]
I think you times all of the inchs together. 5x3x2=30.

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3 0
3 years ago
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an
sleet_krkn [62]

Answer:

-6re−r [sin(6θ) - cos(6θ)]

Step-by-step explanation:

the Jacobian is ∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ

x = e−r sin(6θ), y = er cos(6θ)

δx/δθ = -6rcos(6θ)e−r sin(6θ), δx/δr = -sin(6θ)e−r sin(6θ)

δy/δθ = -6rsin(6θ)er cos(6θ), δy/δr = cos(6θ)er cos(6θ)

∂(x, y) /∂(r, θ) =  δx/δθ × δy/δr - δx/δr × δy/δθ

= -6rcos(6θ)e−r sin(6θ) × cos(6θ)er cos(6θ) - [-sin(6θ)e−r sin(6θ) × -6rsin(6θ)er cos(6θ)]

= -6rcos²(6θ)e−r (sin(6θ) - cos(6θ)) - 6rsin²(6θ)e−r (sin(6θ) - cos(6θ))

= -6re−r (sin(6θ) - cos(6θ)) [cos²(6θ) + sin²(6θ)]

= -6re−r [sin(6θ) - cos(6θ)]     since  [cos²(6θ) + sin²(6θ)] = 1

6 0
3 years ago
Anyone help???????????
Marta_Voda [28]

Answer:

Graph A

hope it helps

Step-by-step explanation:

8 0
3 years ago
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