Answer:
Correct answer: c = 29 m, a = 21 m and b = 20 m
Step-by-step explanation:
Given:
a = b + 1 longer leg a
c = b + 9 hypotenuse c
To solve this problem, we will use Pythagoras' theorem
c² = a² + b²
(b + 9)² = (b + 1)² + b²
b² + 18 b + 81 = b² + 2 b + 1 + b²
2 b² - b² + 2 b - 18 b + 1 - 81 = 0
b² - 16 b - 80 = 0 we will factorize this equation
⇒ b² - 20 b + 4 b - 80 = b (b - 20) + 4 ( b - 20) = (b - 20) · (b + 4) = 0
(b - 20) · (b + 4) = 0 ⇒ b = 20 or b = - 4 false
we accepted b = 20 m
c = b + 9 = 20 + 9 = 29 m
c = 29 m
a = b + 1 = 20 + 1 = 21 m
a = 21 m
God is with you!!!
Answer:
C 33.3333333333
Step-by-step explanation:
Wat y'all must don't no what it is. what grade is y'all in but I need help in some of my stuff
Answer:
The answer is 31 pennies
Step-by-step explanation:
we have to find a number that gives a remainder of 1 when divided by both 2, 3 and 5.
The easiest way to do this is to list the factors of 5, add 1 to it, and test them until we find the one with a remainder of 1 by all three divisors. This is done as follows:
1) factor: 5 (+1) = 6
6 ÷ 2 = 3 remainder 0
6 ÷ 3 = 2 remainder 0
since there is a remainder of 0, when divided by 3, 6 is wrong
2) factor; 10 (+1) = 11
11 ÷ 3 = 3 remainder 2 ( 11 is not the correct answer
3) factor : 15 (+1) = 16
16 ÷ 2 = 8 remainder 0 ( 16 is not the correct answer
4) factor : 20(+1) = 21
21 ÷ 3 = 7 remainder 0 ( 21 is wrong)
5) factor : 25(+1) = 26
26 ÷ 2 = 13 remainder 0 ( 26 is wrong)
6) factor : 30(+1) = 31
31 ÷ 2 = 15 Remainder 1
31 ÷ 3 = 10 Remainder 1
31 ÷ 5 = 6 Remainder 1
since all three divisors give a remainder of 1, the correct answer is 31
Therefore you have 31 pwnnies
Answer:40% weight increase
Step-by-step explanation:
