Answer:
The measure of ∠G is 59°
Step-by-step explanation:
<em>When </em><em>two secants</em><em>, intersect at </em><em>a point outside a circle</em><em> then the </em><em>measure of the angle formed between them</em><em> is </em><em>one-half the positive difference of the measures of the intercepted arcs.</em>
<em></em>
In the given figure
∵ GT is a secant that intersects the circle at points H and T
∵ GE is a secant that intersects the circle at points F and E
∴ The intercepted arcs are HF and TE
∵ GT ∩ GE at G
∵ Point G is outside the circle
→ By using the rule above
∴ m∠G = 
(m arc TE - m arc HF)
∵ m arc TE = 175°
∵ m arc HF = 57°
→ Substitute them in the rule above
∵ m∠G = (175 - 57) = (118)
∴ m∠G = 59
∴ The measure of ∠G is 59°
Answers:
x = 7%
y = 2 slips
Explanation:
The expected value is the result of the sum of each value times its probabilities:
Expeted value = probability 1 × value 1 + probability 2 × value 2 + probability 3 × value 3 + .....
Case 1: at the beginning of the contest:
total number of slips: 30 + 15 + 5 = 50
probability 1 = 30/50
value 1 = 5%
probability 2 = (15/50)
value 2 = x%
probability 3 = (5/50)
value 3 = 15%
⇒ Expected value = 6.6% = (30/50) 5% + (15/50)x% + (5/50)15%
⇒ (15/50)x% = 6.6% - (30/50)5% - (5/50)15%
⇒ (15/50) x% = 2.1%
⇒ x% = (50 / 15) 2.1% = 7%
Answer: 7%
2) Case 2: at one point, ...
Yet, the equation for the expected value is:
Expeted value = probability 1 × value 1 + probability 2 × value 2 + probability 3 × value 3 + .....
Only the probabilities have changed, but the discounts are the same. This is x% is the same value found above: 7%.
The total number of slips now is 4 + y + 2 = 6 + y
And the expected value becomes:
8% = [ 4 / (6+y) ] 5% + [ y / (6 + y) ] 7% + [2 / (6 + y)] 15%
From which you obtain:
Mulitplying by 6+ y: 8% [6 + y] = 4×5% + y×7% + 2×15%
⇒ 8% y + 8%×6 = 4×5% + y 7% + 2×15%
⇒ 8% y - 7%y = 4×5% + 2×15% - 6×8%
⇒ 0.01y = 0.2 + 0.3 - 0.48 = 0.02
⇒ y = 2
Answer: The gravity's constant value is 
.
Explanation:
The given function shows the distance from the ground (in meters) at time t (in sec).
First derivative of displacement is velocity and the second derivative of displacement is acceleration.
Differentiate the given equation with respect to t.
Differentiate the above equation with respect to t.
It is given that gravity is the only acceleration affecting the object and the negative sign shows the downward direction of the object.
Therefore, the gravity's constant value is .
Answer:
3a. 0.415
b. 18.75
Step-by-step explanation:
3a. 2.49/6
=0.415
b. 150/8
=18.75
Answer:
<u>False</u>: C. Completing the square can be used to solve the given equation.
Step-by-step explanation:
First of all, no equation is given. Any answer that suggests a way of finding solutions to the given equation will be false.
The statement of B is debatable. The method of completing the square is most often used for quadratics, but might reasonably be applied in any situation where a perfect square can be created by regrouping the terms.
Statements A and D essentially say the same thing. Both are true. That fact makes statement C false. Statement C is the one you're looking for.
