We know that , in a circle radius perpendicular to chord will bisect the chord.
OM=18, so OQ=QM=18/2=9.
Given QU=8
from figure OQU is a right angled triangle , so OU^2=OQ^2 + QU^2
OU^2 = 9*9 + 8*8 = 81+72=153;
OU=sqrt(153) = 12.37 =13(approx);
From given statements of congruent NT and OU will also be congruent or identical. So, NT=OU=13
M+4n=8
m=n-2
we see that m=n-2
so we can subsitute 'n-2' for m in the first equation
(n-2)+4n=8
n-2+4n=8
5n-2=8
add 2 to both sides
5n=10
divide by 5
n=2
subsitute
n=2
m=n-2
m=2-2
m=0
m=0
n=2
the answer is c. 0,2
First set up the equation to find answer: y=7-x. Then add it to the variable. 3x-1(7-x)=1. So that makes it 3x+-7+-1=1. Add like terms -7+-1=-8, so 3x+-8=1. Now to find x just switch the 8 to the other side. So you end up with 3x=9 since when you take away a number from one side you need to use the opposite value so adding 8 to -8 will cancel it out, then add it to the other, then just divide. 9/3=3, from here just substitute to find y. y=7+3. y=10.
The answer is x=2 and y=5. I hope I helped!
Answer:
swimming pool
Step-by-step explanation: