Answer: Are you ok?
Step-by-step explanation:
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
Your Answer Is approximately 0.143
Answer:
Local minimum at x = 0.
Step-by-step explanation:
Local minimums occur when g'(x) = 0 and g"(x) > 0.
Local maximums occur when g'(x) = 0 and g"(x) < 0.
Set g'(x) equal to 0 and solve:
0 = 2x (x − 1)² (x + 1)²
x = 0, 1, or -1
Evaluate g"(x) at each point:
g"(0) = 2
g"(1) = 0
g"(-1) = 0
There is a local minimum at x = 0.
Answer:
Part 1) 
Part 2) 
Part 3) 
Part 4) 
Step-by-step explanation:
step 1
Find the value of x
In the large right triangle
----> by CAH (adjacent side divided by the hypotenuse)
Remember that

substitute

solve for x
---> improper fraction
step 2
Find the value of z
In the large right triangle
Applying the Pythagorean Theorem

substitute the value of x

solve for z




simplify

step 3
Find the value of y
In the right triangle of the right
---> by SOH (opposite side divided by the hypotenuse)
substitute the given values of y and z
Remember that

so

solve for y


step 4
Find the value of b
In the right triangle of the right
---> by CAH (adjacent side divided by the hypotenuse)
substitute the given values of y and z
Remember that

so

solve for y

