Answer:
sin
(
x/
2
) = -
√
3
/2
Take the inverse sine of both sides of the equation to extract x
from inside the sine.
x/
2
=
arcsin
(
−
√
3/
2
)
The exact value of arcsin
(
−
√
3
/2
) is −
π
/3
.
/x
2
=
−
π
/3
Multiply both sides of the equation by 2
.
2
⋅
x
/2
=
2
⋅
(
−
π
/3
)
Simplify both sides of the equation.
x
=
−
2
π
/3
The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from 2
π
, to find a reference angle. Next, add this reference angle to π to find the solution in the third quadrant.
x
/2
=
2
π
+
π/
3
+
π
Simplify the expression to find the second solution.
x
=
2
π
/3
4
π
Add 4
π to every negative angle to get positive angles.
x
=
10
π
/3
The period of the sin
(
x
/2
) function is 4
π so values will repeat every 4
π radians in both directions.
x
=2
π
/3
+
4
π
n
,
10
π/
3
+
4
π
n
, for any integer n
Exclude the solutions that do not make sin
(
x
/2
)
=
−
√
3/
2 true.
x
=
10
π
/3
+
4
π
n
, for any integer n
Answer:
( x+1)^2 + (y+2)^2 = 25
The center is (-1, -2) and the radius is 5
Step-by-step explanation:
x^2 + 2x + y^2 + 4y = 20
Complete the square
2/2 =1 1^2 =1 so add 1 for x 4/2 =2 2^2 = 4 so add 4 for y
x^2 +2x +1 +y^2 +4y+4 = 20 +1 +4
( x+1)^2 + (y+2)^2 = 25
(x+1) ^2 + (y+2)^2 = 5^2
(x - -1) ^2 + (y - -2)^2 = 5^2
This is in (x-h)^2 + (y-k)^2 = r^2 form where (h,k) is the center and r is the radius
The center is (-1, -2) and the radius is 5
I may be wrong but I hope I am not. But one expression is 16 × .15 and another is 16 × 15/100. When multiplying percentages you have to move the decimal that is invisbly behind the last number, two places to the left. The answer you get is the percentage of the price. You would then take the answer and add it to the price. The answer to the multiplication problem is 2.40. You then add the 2.40 to the $16 and that's the new price.
Answer:
B
Step-by-step explanation:
All functions have y with a power of 1.
option B. is the only one where y has a power of 1.
Answer:
The radius of a sphere is 2 millimeters
The surface area of a sphere is 50.24 square millimeters.
The circumference of the great circle of a sphere is 12.56 millimeters.
Step-by-step explanation:
<u><em>Verify each statement</em></u>
case A) The radius of a sphere is 8 millimeters
The statement is false
we know that
The volume of the sphere is equal to
we have
substitute and solve for r
case B) The radius of a sphere is 2 millimeters
The statement is True
(see the case A)
case C) The circumference of the great circle of a sphere is 9.42 square millimeters
The statement is false
The units of the circumference is millimeters not square millimeters
The circumference is equal to
we have
substitute
case D) The surface area of a sphere is 50.24 square millimeters.
The statement is True
Because
The surface area of the sphere is equal to
we have
substitute
case E) The circumference of the great circle of a sphere is 12.56 millimeters.
The statement is true
see the case C
case F) The surface area of a sphere is 25.12 square millimeters
The statement is false
because the surface area of the sphere is
see the case D
