# ALGEBRA 2 HOMEWORK QUESTION HELP!? Topic: ALGEBRA 2 HOMEWORK QUESTION HELP!?
July 16, 2019 / By Brogan
Question: i have one problem that i dont know how to do for my algebra 2/trig homework. If you could answer this question and try to EXPLAIN how how did it. Thanks in advance. 1.) The hypotenuse of a right triangle is 8.312 cm long. The sum of the lengths of the legs is 10.23 cm. Use a calculator to find the lengths of the legs. Please and thanks. ## Best Answers: ALGEBRA 2 HOMEWORK QUESTION HELP!? Alison | 8 days ago
8.312^2 = x^2 + (10.32-x)^2 69.09 = x^2 + 106.5 - 20.64x + x^2 0 = 2x^2 - 20.64x + 37.41 0 = x^2 - 10.32x + 18.7 x = [10.32 +/- *sqrt(10.32^2 - 4(1)(18.7))] / [2x1] ....
👍 244 | 👎 8
Did you like the answer? ALGEBRA 2 HOMEWORK QUESTION HELP!? Share with your friends Originally Answered: Math homework help( Algebra) another question?
6W+8X=273 Since her total pay was \$273 during the week, this equation is correct (6W (amount on weekdays) 8X amount on weekends)) W+X=42 This is the total number of hours she worked in total. Now it's a matter of substitution. W=42-X 6W+8X=273 6[42-X]+8X=273 252-6X+8X=273 252+2X=273 2X=273-252 2X=21 X=10.5 Hours On the weekends, she worked for 10 and a half hours. Now with this information, sub it in to get the other one. W=42-X W=42-[10.5] W=31.5 So, on the weekdays, she worked for a total of 31 and a half hours. Now to check, 6W+8X=273 6[31.5]+8[10.5]=273 189+84=273 273=273 Correct Ur
After trying to solve problem after problem on yahooanswers, I honestly think most people dont learn a lot if I give them the answers. Heres a VERY similar problem. Just fill in the values from your own problem. (Hint: you will need to use the Quadratic Formula at the end!) Goodluck!!! QUESTION: The hypotenuse of a right triangle has a length of 13 cm. The sum of the lengths of the two legs is 17 cm. Find the lengths of the legs. Let a = length of one leg Let b = length of other leg We are told that: a + b = 10.23 --> b = 10.23 - b  The hypotenuse is 8.312. From Pythagorus: a² + b² = 8.312²  Substitute  into : a² + (8.312 - b)² = 169 This simplifies to: 2a² - 34a + 120 = 0 Divide by 2: a² - 17a + 60 = 0 which factors: (a - 5)(a - 12) = 0 and has roots: a = 5, 12 Substitute into  and get: b = 12, 5 The lengths of the legs are 5 and 12 cm.
👍 100 | 👎 2 Robby
Trey, give me a moment while I grab my abacus, he-he. Just kidding. This is a tricky one. You must've pulled a prank on your teacher to have something like this assigned. Well if you have it you have it, so ... You know that the legs, a and b, sum to 10.23, so a + b = 10.23. You also know that a^2 + b^2 = 8.312^2 since it's a right triangle. Two equations, two unknowns, have at it son.
👍 93 | 👎 -4 Micaiah
isn't that 10.23 divided by 2? sine its the sum of the two legs i'm thinking you need to divide the sum by 2 to get the two lengths of the legs. 10.23 divided by 2 is 5.115 you can round. Pothygreum Therum is A2+B2=C2 (the 2's are exponents) hope this helps.
👍 86 | 👎 -10 Josse
[first, for the sake of clarity, switch f(x) with y] y=5x+4 [then, because this is how you find inverses, switch the (x)s and the (y)s] x=5y+4 [now solve for, or "isolate" y] x-4=5y (x-4)/5=y [so there's your answer, but I think inverse notation looks like this] f^-1(x)=(x-4)/5 [alternatively,] f^-1(x)=x/5+0.8 [or] f^-1(x)=x/5+4/5
👍 79 | 👎 -16 Originally Answered: Pre- Algebra homework question. 30 points?
19c+31=26c-74 +74 -19c Isolate the variable on one side. 105=7c Divide each side by 7. c=15 3/8y-9=13+1/8 y -1/8y +9 Isolate the variable 4(1/4y) = 4(21) Multiply both sides by the reciprocal of the fraction. y = 84 10a-37=6a+51 Isolate the variable. 4a=88 Divide each side by 4. a=22 5w+9.9=4.8+8w Isolate the variable. 5.1=3w Divide each side by 3. w=1.7

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