A community for students.
Here's the question you clicked on:
 0 viewing
lawls
 3 years ago
If the sides of a right triangle have lengths of x7, x an x+1, then x=
(a) (1, 7)
(b) (7, 1)
(c) (7, 4)
(d) (7, 12)
(e) (4, 12)
lawls
 3 years ago
If the sides of a right triangle have lengths of x7, x an x+1, then x= (a) (1, 7) (b) (7, 1) (c) (7, 4) (d) (7, 12) (e) (4, 12)

This Question is Closed

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.0first tell me which is the longest side

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.0x7,x,x+1....... what do u thik have the biggest value

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Hint: Work with the dearest Pythagorean Theorem.

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.0But beofre using Pythagorean Theorem.. u must find the longest side

Calcmathlete
 3 years ago
Best ResponseYou've already chosen the best response.2Obviously, x + 1 would be the largest side and therefore the hypotenuse because of the + 1 while x is neutral and 7 is negative. THerefore, now use Pythagorean THeorem. \[c^2 = a^2 + b^2\]\[(x + 1)^2 = x^2 + (x  7)^2\]\[x^2 + 2x + 1 = x^2 + x^2  14x + 49\]\[x^2 + 2x + 1 = 2x^2  14x + 49\]\[0 = x^2  16x + 48\]Can you figure out how to solve the4 quadratic from here?

lawls
 3 years ago
Best ResponseYou've already chosen the best response.0so answer choice (e) is right?

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0You, sir, are totally correct!

Calcmathlete
 3 years ago
Best ResponseYou've already chosen the best response.2Kinda ironic since 12 is the only true solution, but whatever works ^_^
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.