Answer:
1/64
Step-by-step explanation:
hope that helps
Answer:
24.8
Step-by-step explanation:
We will have to use trigonometry here to find the value of x. The first thing we can see is that we need to work out the hypotenuse. We need to chose whether to use Sin, Cos or Tan. We do that by look at the sides .
→ 'x' is the hypotenuse
→ 12 is opposite
→ The adjacent is the side that doesn't have a numerical value to we look for a formula triangle without adjacent in it
Tan = Opposite ÷ Adjacent
Sin = Opposite ÷ Hypotenuse
Cos = Adjacent ÷ Hypotenuse
We can see the Sin formula has no adjacent in it so we use that formula. However we want to work out the hypotenuse so we have to rearrange the sin formula to get hypotenuse as the subject so
Sin = Opposite ÷ Hypotenuse
Hypotenuse = Opposite ÷ Sin
→ Let's substitute in the values
Hypotenuse = 12 ÷ Sin(29)
Hypotenuse = 24.7519841
So the value of x is 24.8 to 1 decimal place
Answer:
24
Step-by-step explanation:
The Pythagorean Theorem states that 
where a and b are the legs in a right triangle and c is the hypotenuse. Therefore, just substitute the values a = 10 and c = 26 into the equation to get b = 24
Hope this helps :)
![\bf \textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2) ~\hfill a^3-b^3 = (a-b)(a^2+ab+b^2) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \boxed{a^6+b^6}\implies a^{2\cdot 3}+b^{2\cdot 3}\implies (a^2)^3+(b^2)^3 \\[2em] [a^2+b^2] [(a^2)^2-a^2b^2+(b^2)^2]\implies \boxed{(a^2+b^2)(a^4-a^2b^2+b^4)}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdifference%20and%20sum%20of%20cubes%7D%20%5C%5C%5C%5C%20a%5E3%2Bb%5E3%20%3D%20%28a%2Bb%29%28a%5E2-ab%2Bb%5E2%29%20~%5Chfill%20a%5E3-b%5E3%20%3D%20%28a-b%29%28a%5E2%2Bab%2Bb%5E2%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cboxed%7Ba%5E6%2Bb%5E6%7D%5Cimplies%20a%5E%7B2%5Ccdot%203%7D%2Bb%5E%7B2%5Ccdot%203%7D%5Cimplies%20%28a%5E2%29%5E3%2B%28b%5E2%29%5E3%20%5C%5C%5B2em%5D%20%5Ba%5E2%2Bb%5E2%5D%20%5B%28a%5E2%29%5E2-a%5E2b%5E2%2B%28b%5E2%29%5E2%5D%5Cimplies%20%5Cboxed%7B%28a%5E2%2Bb%5E2%29%28a%5E4-a%5E2b%5E2%2Bb%5E4%29%7D)
about the second one... well, is a "fait accompli" that using the pythagorean theorem, if x = 8 and y = 5, the hypotenuse must be √(8² + 5²) = √(89), which is neither of those choices.
5, 8, 13 are no dice, namely 5² + 8² ≠ 13
25, 64, 17 is are no dice too, because 25² + 17² ≠ 64²
however, 5,12 and 13 are indeed a pythagorean triple
also is 39, 80, 89.
when looking for a pythagorean triple, recall that c² = a² + b².
so the longest leg is the sum of the square of the small ones.
so what you'd do is, check the small legs, square them, add them up, if they're indeed a pythagorean triple, they "must" add up to the longest leg.
3 1/4 times 4 is 16 then time 5 1/2 =71.5
