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fomenos
2 years ago
12

3(x-6)^4+11=15 solve for x

Mathematics
1 answer:
Tamiku [17]2 years ago
7 0

Answer:

x = 7.07 or 4.925

Step-by-step explanation:

3(x-6)^4+11=15

3(x-6)^4=15-11

(x-6)^4=4/3

x-6 = +\sqrt[4]{4/3} or -\sqrt[4]{4/3}

x= +\sqrt[4]{4/3} +6 or -\sqrt[4]{4/3}  +6

x = 7.07 or 4.925

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Question is Incomplete, Complete question is given below.

Prove that a triangle with the sides (a − 1) cm, 2√a cm and (a + 1) cm is a right angled triangle.

Answer:

∆ABC is right angled triangle with right angle at B.

Step-by-step explanation:

Given : Triangle having sides (a - 1) cm, 2√a and (a + 1) cm.  

We need to prove that triangle is the right angled triangle.

Let the triangle be denoted by Δ ABC with side as;

AB = (a - 1) cm

BC = (2√ a) cm

CA = (a + 1) cm  

Hence,

{AB}^2 = (a -1)^2 

Now We know that

(a- b)^2 = a^2+b^2 - 2ab

So;

{AB}^2= a^2 + 1^2 -2\times a \times1

{AB}^2 = a^2 + 1 -2a

Now;

{BC}^2 = (2\sqrt{a})^2= 4a

Also;

{CA}^2 = (a + 1)^2

Now We know that

(a+ b)^2 = a^2+b^2 + 2ab

{CA}^2= a^2 + 1^2 +2\times a \times1

{CA}^2 = a^2 + 1 +2a

{CA}^2 = AB^2 + BC^2

[By Pythagoras theorem]

a^2 + 1 +2a = a^2 + 1 - 2a + 4a\\\\a^2 + 1 +2a= a^2 + 1 +2a

Hence, {CA}^2 = AB^2 + BC^2

Now In right angled triangle the sum of square of two sides of triangle is equal to square of the third side.

This proves that ∆ABC is right angled triangle with right angle at B.

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Answer:

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Step-by-step explanation:

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The appropriate choice is
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