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
The asymptote cannot be x= because x can be any number. If you think about it, you can take a number to any exponent.
If x is a positive exponent, y is positive.
If x is a nevative exponent, y decreases, but is still positive. This is because a number to a negative exponent equals 1 over the number to the positive exponent. Thus, it is smaller, but still positive.
If x is 0, y is positive again because anything to the zero is positive 1.
There is no way y could be less than or equal to zero. So, there is an asymptote at y=0.
Also, set the equation equal to 0 and solve. You should end up with 4^x=0. Since no exponenent can make a number zero, this isn't possible, so y cannot equal zero.
Here is the graph for a visual:
Answer:
Yes, it's 13r.
Step-by-step explanation:
5r+8r=13r
Answer:
Remove parentheses. xy³
z^4
4Step-by-step explanation: