Answer:
Step-by-step explanation:
4.
a)
P = 2width + 2length
8x = 2x + 2 length; 8x-2x = 2 length; 6x=2 length
length = 3x
b)
sides= 1/4 of 3x = 3x/4
perimeter of square = 4* side = 4*3x/4 = 3x
c)
x= 4
P rectangle =8x= 8*4= 32
P square = 3x = 3*4 = 12
32-12 = 20 cm greater is the P rectangle than P of square
d)
A rectangle = 2x*x = 2*4*4 = 32
A square =(3x/4)² = (3*4/4)² = 9
32-9 = 23 cm² greater is the A rectangle than A of square
5.
4Comic+1 Nonfiction
4comic = 8y so 1 comic = 2y
1 nonfiction = (3+7y)+ 2y
a)
1 nonfiction = 7y+2y+3 = 9y+3
b)
total= 8y+9y+3 = 17y+3
if y=4
total = 17*4 +3 = $71
Answer:
Step-by-step explanation:
![\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.
Answer:
22:35
Step-by-step explanation:
Find the GCD (or HCF) of numerator and denominator
GCD of 66 and 105 is 3
Divide both the numerator and denominator by the GCD
66 ÷ 3/105 ÷ 3
Reduced fraction:
22/35
therefore the ratio will be 22:35
Answer:
(6,5), (3,-4) either one of those...
