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pychu [463]
3 years ago
13

A right-angled triangle has shorter side lengths exactly c^2-b^2 and 2bc units respectively, where b and c are positive real num

bers such that cc is greater than b. Find an exact expression for the length of the hypotenuse (in appropriate units).
Mathematics
1 answer:
yKpoI14uk [10]3 years ago
3 0

Answer: hypotenuse = c^{2} + b^{2}

Step-by-step explanation: Pythagorean theorem states that square of hypotenuse (h) equals the sum of squares of each side (s_{1},s_{2}) of the right triangle, .i.e.:

h^{2} = s_{1}^{2} + s_{2}^{2}

In this question:

s_{1} = c^{2}-b^{2}

s_{2} = 2bc

Substituing and taking square root to find hypotenuse:

h=\sqrt{(c^{2}-b^{2})^{2}+(2bc)^{2}}

Calculating:

h=\sqrt{c^{4}+b^{4}-2b^{2}c^{2}+(4b^{2}c^{2})}

h=\sqrt{c^{4}+b^{4}+2b^{2}c^{2}}

c^{4}+b^{4}+2b^{2}c^{2} = (c^{2}+b^{2})^{2}, then:

h=\sqrt{(c^{2}+b^{2})^{2}}

h=(c^{2}+b^{2})

Hypotenuse for the right-angled triangle is h=(c^{2}+b^{2}) units

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What is the cubed root of 27?
lianna [129]

Answer:

Then answer is 3

Step-by-step explanation:

cube root = ³√27

3×3×3 = 3³ = 27

6 0
3 years ago
Read 2 more answers
Solve the equation for the indicated variable S=2πrh+2πr²<br> Solve for h
valina [46]

Answer:

Good luck :)

Step-by-step explanation:

S=2πrh+2πr ²

subtract 2πr ² from each side

S -2πr ² = 2πrh

divide by 2πr from each side

(S -2πr ² )/ 2πr  = h

4 0
3 years ago
The set of the consecutive odd numbers 1, 3, 5, 7, ... , N has a sum of 400. How many numbers are in the set?
Amanda [17]

Well, we could try adding up odd numbers, and look to see when we reach 400. But I'm hoping to find an easier way.

First of all ... I'm not sure this will help, but let's stop and notice it anyway ...
An odd number of odd numbers (like 1, 3, 5) add up to an odd number, but
an even number of odd numbers (like 1,3,5,7) add up to an even number.
So if the sum is going to be exactly 400, then there will have to be an even
number of items in the set.

Now, let's put down an even number of odd numbers to work with,and see
what we can notice about them:

         1, 3, 5, 7, 9, 11, 13, 15 .

Number of items in the set . . . 8
Sum of all the items in the set . . . 64

Hmmm.  That's interesting.  64 happens to be the square of 8 . 
Do you think that might be all there is to it ?

Let's check it out:

Even-numbered lists of odd numbers:

1, 3                                   Items = 2, Sum = 4
1, 3, 5, 7                           Items = 4, Sum = 16
1, 3, 5, 7, 9, 11                 Items = 6, Sum = 36
1, 3, 5, 7, 9, 11, 13, 15 . . Items = 8, Sum = 64 .

Amazing !  The sum is always the square of the number of items in the set !

For a sum of 400 ... which just happens to be the square of 20,
we just need the <em><u>first 20 consecutive odd numbers</u></em>.

I slogged through it on my calculator, and it's true.

I never knew this before.  It seems to be something valuable
to keep in my tool-box (and cherish always).


3 0
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Please help pic below
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How much is it but don’t stop it fast cash x
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Answer:

yes

Step-by-step explanation:

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