A² + b² = c²
30² + 72² = c²
<u />900 + 5,184 = c²
6084 = c²
(Square root both sides)
c = 78 m
<u />
First Let we solve the Original system of equations:
equation (1): 
equation (2): 
Multiplying equation (1) by 7, we get
Subtracting,
implies 
Then
Thus the solution of the original equation is
Now Let we form the new equation:
Equation 2 is kept unchanged:
Equation (2):
Equation 1 is replaced with the sum of equation 1 and a multiple of equation 2:
Equation (1): 
Now solve this two equations: 
Multiply (1) by 7 and (2) by 8,
Subtracting,
implies
Then x=2.
so the solution for the new system of equation is x=2, y=1.
This Show that the solution to the system of equations 8x − 5y = 11 and 7x − 8y = 6 is the same as the solution to the given system of equations
Answer:
5/14
Step-by-step explanation:
I assume after testing the 1st car, it is not placed back into the pool.
So, 1st test orange is 5/8
2nd test orange is 4/7.
Both had to be true, so 5/8 x 4/7 = 5/14
Compatible numbers are numbers that are close to the numbers they're replacing that divide evenly into each other
NOTES<span><span> Compatible numbers are numbers that are close to the numbers they're replacing that divide evenly into each other</span><span> <span>When we estimate we're trying to make the problem easier, so we can solve it without doing calculationsDividing is easier when we have numbers that divide evenly into each other, so that's how we'll estimate</span></span><span> <span>56,000 is like 560 times 100800 is like 8 times 100When we divide 56,000 by 800, we can cancel out the 100's, and we're left with 560 divided by 8In other words, since there are two zeroes in the divisor, we cancel out two zeros in the dividend</span></span><span> Since we rounded the dividend and divisor into such nice numbers, we can easily divide the front digits to get a round number!</span><span> Since we canceled a few zeroes, there's only one zero left</span><span> </span><span> <span>There are usually more than just one pair of compatible numbers to choose fromAs long as the numbers are easily divisible and close to the original numbers, they'll be good compatible numbers!<span>We'll get a slightly different answer, but it's still a good estimate of our quotient</span></span></span></span>
