Step-by-step explanation:
Firstly, we have to find m∠J.
Since all the angles of a Δ equal 180°, angles J, L, and K should have a sum of 180°.
So,
m∠J + m∠L + m∠K = 180°
The diagram shows us that ∠L = 49° and ∠K = 90°, so we plug in those numbers in the equation.
m∠J + 49° + 90° = 180°
Then we simplify
m∠J + 139° = 180°
Subtract 139° to both sides
∠J = 41
Now the other angles.
Since ΔJKL ~ ΔRST, then ∠J ≅ ∠R, ∠K ≅ ∠S, and ∠L ≅ ∠T
Meaning, m∠J = m∠R, m∠K = m∠S, and m∠L = m∠T
Since we know m∠J = 41°, m∠K = 90°, and m∠L = 49° we could plug those in so...
41° = m∠R , 90° = m∠S , and 49° = m∠T
Answer:
Given: The radius of circle C is 6 units and the measure of central angle ACB is StartFraction pi Over 2 EndFraction radians.
What is the approximate area of the entire circle?
113 square units
What is the approximate area of the entire sector created by central angle ACB?
28 square units
What is the approximate area of the shaded region only?
22 square units
400 should be the answer. Hope this helps!
Answer is:
It includes points in quadrant II and it doesnt include points in quadrant I
Explanation:
For odd functions a rule is:
f(x) = -f(-x) or in other words
f(x) + f(-x) = 0
Because of this function can be in quadrants I and III or in quadrants II and IV as in pairs...
1) m=2
B=8
Y=2x+8
2) the slop is 2
3)the slope is 0
4)the slop is -4/3
