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Nana76 [90]
3 years ago
11

Find x! Please help!!

Mathematics
1 answer:
a_sh-v [17]3 years ago
5 0

{(8 + x)}^{2}  =  {8}^{2}  +  {15}^{2}  \\ {(8 + x)}^{2}  =  64  +  225 \\  {(8 + x)}^{2}  = 289 \\ {(8 + x)}^{2}  =  {17}^{2}  \\ 8 + x = 17 \\ so \: x = 17 - 8 = 9
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Is X-1 a factor of x^5-1?
Musya8 [376]

Answer: No

Explanation:

According to factor theorem, if f(x)=0 then x is a factor of the given function or equation.

As x-1 is a factor

We equate x-1=0

x=1

Substituting in x^5-1, we have 1^5-1 =1-1=0.

Hence, it's a factor.

When coming to x^5+1, it would become 1^5+1=1+1=2

So x-1 isn't a factor of x^5+1.

4 0
2 years ago
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2 years ago
Arc MNP equals 126. O is the center. Find angle MOP
Paha777 [63]
We know that

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3 0
3 years ago
Prove that root 7 is irrational by the method of contradiction
Alchen [17]

Let assume that \sqrt7 is a rational number. Therefore it can be expressed as a fraction \dfrac{a}{b} wherea,b\in\mathbb{Z} and \text{gcd}(a,b)=1.

\sqrt7=\dfrac{a}{b}\\\\7=\dfrac{a^2}{b^2}\\\\a^2=7b^2

This means that a^2 is divisible by 7, and therefore also a is divisible by 7.

So, a=7k where k\in\mathbb{Z}

(7k)^2=7b^2\\\\49k^2=7b^2\\\\7k^2=b^2

Analogically to a^2=7b^2 ------- b^2 is divisible by 7 and therefore so is b.

But if both numbers a and b are divisible by 7, then \text{gcd}(a,b)=7 which contradicts our earlier assumption that \text{gcd}(a,b)=1.

Therefore \sqrt7 is an irrational number.

8 0
4 years ago
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