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Anton [14]
3 years ago
5

Can someone help pls

Mathematics
2 answers:
Roman55 [17]3 years ago
8 0

Answer:

That means the missing side length is 15

Step-by-step explanation:

We use the Pythagorean theorem to find the hypotenuse. The hypotenuse for this triangle is 17. The equation for the Pythagorean theorem is a^2 + b^2 = c^2. Our equation is x^2+64 = 289. We can now subtract 64 from both sides and get 225. The square root of 225 will be 15.

balandron [24]3 years ago
8 0

Answer:

hi

Step-by-step explanation:

{c}^{2}  =  {a}^{2}  + {b}^{2}  \\  {8}^{2}  +  {b}^{2}  =  {17}^{2}  \\ {b}^{2}  =  {17}^{2}  -  {8}^{2}  \\  {b}^{2}  = 289 - 64 \\  {b}^{2}  = 225 \\ b =  \sqrt{225}  \\ b = 15

hope it helps

Have a nice day

byee

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Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .
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Answer:

The family of all prime numbers such that a^{2} + b^{2} + c^{2} -1 is a perfect square is represented by the following solution:

a is an arbitrary prime number. (1)

b = \sqrt{1 + 2\cdot a \cdot c} (2)

c is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if (x+y)^{2} = x^{2} + 2\cdot x\cdot y  + y^{2}. From statement, we must fulfill the following identity:

a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}

By Associative and Commutative properties, we can reorganize the expression as follows:

a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2} (1)

Then, we have the following system of equations:

x = a (2)

(b^{2}-1) = 2\cdot x\cdot y (3)

y = c (4)

By (2) and (4) in (3), we have the following expression:

(b^{2} - 1) = 2\cdot a \cdot c

b^{2} = 1 + 2\cdot a \cdot c

b = \sqrt{1 + 2\cdot a\cdot c}

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, a, b, c > 1. If a, b and c are prime numbers, then  2\cdot a\cdot c must be an even composite number, which means that a and c can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.

In addition, b must be a natural number, which means that:

1 + 2\cdot a\cdot c \ge 4

2\cdot a \cdot c \ge 3

a\cdot c \ge \frac{3}{2}

But the lowest possible product made by two prime numbers is 2^{2} = 4. Hence, a\cdot c \ge 4.

The family of all prime numbers such that a^{2} + b^{2} + c^{2} -1 is a perfect square is represented by the following solution:

a is an arbitrary prime number. (1)

b = \sqrt{1 + 2\cdot a \cdot c} (2)

c is another arbitrary prime number. (3)

Example: a = 2, c = 2

b = \sqrt{1 + 2\cdot (2)\cdot (2)}

b = 3

4 0
3 years ago
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Svetlanka [38]

Answer:

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Step-by-step explanation:

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Dominik [7]

Answer:

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Solution :

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Step-by-step explanation:

Given

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