The solutions appear to be
{π/2, 2π/3, 4π/3}.
_____
Replacing sin(2x) with 2sin(x)cos(x), you have
2sin(x)cos(x) +sin(x) -2cos(x) -1 = 0
sin(x)(2cos(x) +1) -(2cos(x) +1) = 0 . . . . factor by grouping
(sin(x) -1)(2cos(x) +1) = 0
This has solutions
sin(x) = 1
x = π/2and
2cos(x) = -1
cos(x) = -1/2
x = {2π/3, 4π/3}
Answer:
Luca bought 2 steaks and 4 watermelons.
Step-by-step explanation:
During the sale, steaks sold for $4 and watermelon sold for $3.50.
Let the number of steaks Luca bought be represented as x and the number of watermelons Luca bought be represented as y.
Luca spent $22 in total.
That means that the number of steaks that Luca bought multiplied by the price of each steaks plus the number of watermelons that Luca bought multiplied by the price of each watermelon will give $22.
In other words:
(4 * x) + (3.5 * y) = 22__________________(1)
Luca bought a total of 6 steaks and watermelons. This means that:
x + y = 6____________________________(2)
We have two simultaneous equations.
4x + 3.5y = 22 ___________(1)
x + y = 6 _______________(2)
From (2):
x = 6 - y
Putting this back in (1):
4(6 - y) + 3.5y = 22
24 - 4y + 3.5y = 22
24 - 0.5y = 22
=> 24 - 22 = 0.5y
2 = 0.5y
=> y = 2/0.5 = 4
Hence:
x = 6 - 4 = 2
Luca bought 2 steaks and 4 watermelons.
Answer:
41.92%
Step-by-step explanation:
In table attached standard normal distribution curve is shown.
Data
mean, µ = 4.3 minutes
standard deviation, σ = 0.5 minutes
desired point, x = 5 minutes
x-axis, Z = (x - µ)/σ
The probability that calls last between 4.3 and 5.0 minutes is equivalent to find the area below the curve between Z = 0 and Z = (5 - 4.3)/0.5 = 1.4, which is 0.4192, or 41.92%
Answer:
Area = 3.3998
Perimeter = 2.9
Step-by-step explanation:
A method for calculating the area of a triangle when you know the lengths of all three sides.
Let a,b,c be the lengths of the sides of a triangle. The area is given by:
Area = √ p ( p − a ) ( p − b ) ( p − c )
where p is half the perimeter, or
a + b + c / 2
p = 1.7 + 1.7 + 2.4 / 2 = 5.8 / 2 = 2.9
a = Area = 3.3998
Heron was one of the great mathematicians of antiquity and came up with this formula sometime in the first century BC, although it may have been known earlier. He also extended it to the area of quadrilaterals and higher-order polygons.
