Answer:

Step-by-step explanation:
we know that
The inscribed angle is half that of the arc it comprises.
so
In this problem
Let
x-----> the measure of an arc intercepted by an inscribed angle


Answer:
$82 sundaes
$74 banana split
Step-by-step explanation:
Step one :
Given data
Cost of sundaes =$2 each
Cost of banana split =$3 each
Let the number of sundaes bought be x
And the number of banana split bought be y
The total is x+y
The shop sold 8 more Sundaes than the Banana split
x=8+y—-------1
Total amount made = $156
Therefore the system of equation for the situation is
x=8+y----—----------1
2x+3y=156---------2
But x=8+y
8+y+y=156
8+2y=156
2y=156-8
2y=148
Divide both sides by 2
y=148/2
y=74
On That day they sold $74 banana split
And also they sold x=8+74=$82 sundaes
Answer:
(a + 2b)(a - b)
Step-by-step explanation:
Assuming you require the expression to be factored
Given
a² +
ab - b² ← factor out
from each term
=
(a² + ab - 2b²) ← factor the quadratic
Consider the factors of the coefficient of the b² term(- 2) which sum to give the coefficient of the ab- term (+ 1)
The factors are + 2 and - 1, since
2 × - 1 = - 2 and 2 - 1 = + 1, thus
a² + ab - 2b² = (a + 2b)(a - b) and
a² +
ab - b² =
(a + 2b)(a - b)
Answer:

Step-by-step explanation:
given,
angular deceleration, α = -0.5 rad/s²
final angular velocity,ω_f = 0 rad/s
angular position, θ = 6.1 rad
angular position at 3.9 s = ?
now, Calculating the initial angular speed




now, angular position calculation at t=3.9 s



Hence, the angular position of the wheel after 3.9 s is equal to 5.83 rad.
What we know:
Football field is 100 yards in length
End zones are 10 yards each in length
Perimeter between pylons is 306 2/3 yards
What we need to find:
a. Perimeter and area of one end zone
b. Perimeter and area of without end zones
c. Perimeter and area of playing field with end zones
First we need to find the measurements of the field between pylons using the perimeter of 306 2/3 yards. We already know the length is 100 yards so we need to find width (w).
P=2l + 2w
306 2/3=2 (100) + 2w
306 2/3= 200 + 2w
300 2/3-200=200-200+2w
106 2/3=2w
106 2/3/2=2/2w
53 1/3=w
a. Perimeter=2 (10)+2 (53 1/3)=126 2/3 yards
Area=10×53 1/3=533 1/3 yd²
b. Perimeter=2 (100)+2 (53 1/3)=306 2/3 yards
Area=100×53 1/3=5333 1/3 yd²
c. Perimeter=2 (120) + 2 (53 1/3)=346 2/3 yards
Area=120×53 1/3=6400 yd²