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
Answer:
Solution
Correct option is C)
(fog)(x)=f(g(x))=2(3x+2)−1=6x+4−1=6x+3=3(2x+1)
Answer:
a) b = 8, c = 13
b) The equation of graph B is y = -x² + 3
Step-by-step explanation:
* Let us talk about the transformation
- If the function f(x) reflected across the x-axis, then the new function g(x) = - f(x)
- If the function f(x) reflected across the y-axis, then the new function g(x) = f(-x)
- If the function f(x) translated horizontally to the right by h units, then the new function g(x) = f(x - h)
- If the function f(x) translated horizontally to the left by h units, then the new function g(x) = f(x + h)
In the given question
∵ y = x² - 3
∵ The graph is translated 4 units to the left
→ That means substitute x by x + 4 as 4th rule above
∴ y = (x + 4)² - 3
→ Solve the bracket to put it in the form of y = ax² + bx + c
∵ (x + 4)² = (x + 4)(x + 4) = (x)(x) + (x)(4) + (4)(x) + (4)(4)
∴ (x + 4)² = x² + 4x + 4x + 16
→ Add the like terms
∴ (x + 4)² = x² + 8x + 16
→ Substitute it in the y above
∴ y = x² + 8x + 16 - 3
→ Add the like terms
∴ y = x² + 8x + 13
∴ b = 8 and c = 13
a) b = 8, c = 13
∵ The graph A is reflected in the x-axis
→ That means y will change to -y as 1st rule above
∴ -y = (x² - 3)
→ Multiply both sides by -1 to make y positive
∴ y = -(x² - 3)
→ Multiply the bracket by the negative sign
∴ y = -x² + 3
b) The equation of graph B is y = -x² + 3
Your answer is 38 I hope its correct :)
