## anonymous one year ago What is the value of x? Round to the nearest tenth.

1. anonymous

2. SolomonZelman

$$\large\color{black}{ \displaystyle a^2+b^2=c^2 }$$ where c is the hypotenuse of a right triangle, and a & b are the legs.

3. SolomonZelman

Would you then agree with the following set up (to find x)? $$\large\color{black}{ \displaystyle 12^2+x^2=22^2 }$$

4. anonymous

Yes

5. SolomonZelman

ok, then tell me what x is going to equal (and disregard the negative x solution)

6. anonymous

10

7. SolomonZelman

no

8. SolomonZelman

it is not 12+x=22 then it would have been correct, BUT///

9. SolomonZelman

it is 12²+x²=22²

10. anonymous

oh ok

11. SolomonZelman

first, you have to simplify the powers.

12. SolomonZelman

12²=? 22²=?

13. anonymous

144

14. anonymous

484

15. SolomonZelman

yes

16. SolomonZelman

$$\large\color{black}{ \displaystyle 12^2+x^2=22^2 }$$ $$\large\color{black}{ \displaystyle 144+x^2=484 }$$ when you subtract 144 from both sides, you get: $$\large\color{black}{ \displaystyle x^2=340 }$$

17. SolomonZelman

So, x² is 340

18. anonymous

okay

19. SolomonZelman

then take the square root of both sides.

20. anonymous

okay

21. anonymous

then?

22. anonymous

so what's the final answer? @SolomonZelman

23. anonymous

@SolomonZelman i running out of time please

24. anonymous

i'm*

25. SolomonZelman

$$\large\color{black}{ \displaystyle x^2=340 }$$ $$\large\color{black}{ \displaystyle \sqrt{x^2}=\sqrt{340} }$$ $$\large\color{black}{ \displaystyle x=\pm \sqrt{340} }$$ exclude the negative result, as a side can't be negagive. $$\large\color{black}{ \displaystyle x=\sqrt{340} }$$ $$\large\color{black}{ \displaystyle x=\sqrt{4\times 85} }$$ $$\large\color{black}{ \displaystyle x=2\sqrt{85} }$$