Which model shows 25%?
1 answer:
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Answer:
FH ~ 10.02
Step-by-step explanation:
1. Approach
One should first find the circumference of the given circle. Then one should find how large the fraction of the circumference one is supposed to find is. Finally, one should multiply the fraction of the circumference one is supposed to find by the total circumference.
2. Circumference of the circle
The formula for circumference is;
π
Substitute in the given values;
It is given that the radius is, hence
2 (7) π
14π
3. Find the fraction of the circumference one is supposed to find
It is given that the angles over the measure of the total degrees of angles in a circle are equal to the arc surrounding the angles of the circumference. Essentially;
Substitute in the given information and solve;
arc =
arc =
arc ~ 10.02
ANSWER
D.
EXPLANATION
The given function is
Comparing to the general form of a trigonometric sine function,
The amplitude is absolute value of A
The period is given by
is a shift to the right of 4 units.
and
is an upward shift of 9 units
Therefore the phase shift is (4,9).
The correct answer is D
D.
EXPLANATION
The given function is
Comparing to the general form of a trigonometric sine function,
The amplitude is absolute value of A
The period is given by
is a shift to the right of 4 units.
and
is an upward shift of 9 units
Therefore the phase shift is (4,9).
The correct answer is D
Answer:
Step-by-step explanation:
9) PQR Is an isosceles triangle
=> ∠PRQ = (180° - x)/2
PRS is an isosceles right triangle
=> ∠PRS = 45°
Have: ∠PRS + ∠PRQ = 115°
=>
=> 180 - x = (115 - 45).2 = 140
<=> x = 180 - 140 = 40
10) ABD is an isosceles right triangle => ∠ABD = 45°
BCD is an equilateral triangle => ∠CBD = 60°
have: x = ∠ABD + ∠CBD = 45° + 60° = 105°
11) have: x = y (2)
PQT is an isosceles triangle => ∠PQT = 180 - 70.2 = 40
QTS is an isosceles triangle => ∠TQS = 180 -2x
QRS is an isosceles triangle => ∠RSQ = y
have: 40 + 180 - 2x + y = 180 => 2x - y = 40 (1)
(1)(2) =>
=> x + y = 80
12) EFJ Is an equilateral triangle => ∠FJE = 60
∠FJE is the outer angle of the triangle FHJ but FHJ is an isosceles triangle
=> 60 = 2.∠JHF => ∠JHF = 30°
∠JHF is the outer angle of the triangle FHG
=> 30° = 2x
<=> x = 15°