A person is on the outer edge of a carousel with a radius of 20 feet that is rotating counterclockwise around a point that is centered at the origin. What is the exact value of the position of the rider after the carousel rotates 5pi/12 radians?
a. {5[sqrt(2)-sqrt(6)], 5[sqrt(2)+sqrt(6)]}
b. {5[sqrt(2)+sqrt(6)], 5[-sqrt(2)+sqrt(6)]}
c. {5[sqrt(2)+sqrt(6)], 5[sqrt(2)-sqrt(6)]}
d. {5[-sqrt(2)+sqrt(6)], 5[sqrt(2)+sqrt(6)]}
Can someone explain how to do this?
Sum and differences formula quick check:
1. D
2. D
3. B
4. B
5. D
Just took it.
I will switch to degrees, sensing that you may be more familiar with those units
5π/12 radians = 75° = 30° + 45°
draw a unit circle and consider any point (x,y) on that circle.
Construct any right-angled triangle with base x and height y and hypotenuse 1
let the angle at the centre be Ø
then sinØ = y/1 -----> y = sinØ
and cosØ = x/1 ----> x = cosØ
If we make the radius r instead of 1, our point becomes
(rcosØ, rsinØ) or (20cos75°, 20sin75°)
so we need both cos75° and sin75°
recall sin(a+b) = sina cosb + cosa sinb
sin(75) = sin(30+45) = sin30cos45 + cos30sin45
= (1/2)(√2/2) + (√3/2)(√2/2) = (√2 + √6)/4
similarly:
cos(75) = cos(30+45)
= cos30cos45 - sin30sin45
= (√3/2)(√2/2) - (1/2)(√2/2)
= (√6 - √2)/4
so your point (x,y)
= (20(√6-√2/4) , 20((√2 + √6)/4)
= (5√6 - 5√2 , 5√2 + 5√6)
looks like d) is our match
stobotnik is right
those answers still 100% right thx stobotnik
thank you !!!
apply the distributive property (2 sqrt x - 1)(5 + sqrt x)
simplify the radical 2 sqrt x-1 * 5+ sqrt x
To simplify the expression (2√x - 1)(5 + √x), you can use the distributive property.
(2√x - 1)(5 + √x) = 2√x * 5 + 2√x * √x - 1 * 5 - 1 * √x
= 10√x + 2x - 5 - √x
= (10√x - √x) + 2x - 5
= 9√x + 2x - 5.
So, the simplified form of (2√x - 1)(5 + √x) is 9√x + 2x - 5.
combine like term 2 sqrt x-1 * 5+ sqrt x
To combine like terms in the expression (2√x - 1)(5 + √x), we need to distribute the terms and then simplify:
(2√x - 1)(5 + √x) = 2√x * 5 + 2√x * √x - 1 * 5 - 1 * √x
= 10√x + 2x - 5 - √x
To combine like terms, we can add together the coefficients of √x:
10√x + 2x - 5 - √x = (10 - 1)√x + 2x - 5
= 9√x + 2x - 5.
So, the expression (2√x - 1)(5 + √x) simplifies to 9√x + 2x - 5 after combining like terms.
state the variable restriction 2 sqrt x-1 * 5+ sqrt x
In the expression (2√x - 1)(5 + √x), there are no variable restrictions. We are free to use any non-negative real number for the variable x.
\frac{8-\sqrt{x^3}}{3\sqrt{x}}
To find the variable restriction for the expression:
\(\frac{8 - \sqrt{x^3}}{3 \sqrt{x}}\)
We need to ensure that the denominator, \(3 \sqrt{x}\), is not equal to zero. Since division by zero is undefined, we must have \(\sqrt{x} \neq 0\).
This means that \(x\) must be greater than zero. Therefore, the variable restriction for the expression is \(x > 0\).
The solution to square root of x minus 2 plus 4 equals 1 is 11.
To verify if the solution to the equation √x - 2 + 4 = 1 is indeed 11, let's solve the equation.
√x - 2 + 4 = 1
First, let's simplify the equation by combining like terms:
√x + 2 = 1
Next, let's isolate the square root term by subtracting 2 from both sides of the equation:
√x = 1 - 2
√x = -1
Since the square root of a number cannot be negative, we conclude that there are no real solutions to the original equation. Therefore, the solution x = 11 is incorrect.
\sqrt{7x-5}
The expression \(\sqrt{7x-5}\) represents the square root of \(7x-5\). It does not represent an equation or a solution; it is just the expression for the square root of \(7x-5\).
\frac{\sqrt{x-5}}{3}
The expression \(\frac{\sqrt{x-5}}{3}\) represents the square root of \(x-5\) divided by 3. It does not represent an equation or a solution; it is just the expression for the square root of \(x-5\) divided by 3.
A water sprinkler that has a circular spray pattern is used to water the grass in the picture below.
The radius of the sprayed water can be represented by the formula r equals square root of 1.63 A end root, where A is the area of the grass watered in square metres.
*Notice that the are of grass watered is not a perfect circle*
If the radius of the sprinkler is 1.45m, the area of grass, to the nearest hundredth of a metre, that will be watered is Answer
m.
To find the area of the grass that will be watered by the sprinkler, we can use the given formula:
r = √(1.63A), where r is the radius of the sprinkler and A is the area of the grass watered in square meters.
We are given that the radius of the sprinkler is 1.45m, so we can substitute this value into the formula:
1.45 = √(1.63A)
To solve for A, we need to isolate it by squaring both sides of the equation:
(1.45)^2 = (√(1.63A))^2
2.1025 = 1.63A
Divide both sides of the equation by 1.63 to solve for A:
A = 2.1025 / 1.63
A ≈ 1.29
Therefore, the area of the grass that will be watered, to the nearest hundredth of a meter, is approximately 1.29 square meters.
a car is traveling at a speed pf 80 km/h on a icy highway. the driver swerves while applying the brakes to avoid a collision. if the Yaw mark that car leaves on the icy asphalt has a radius of 270m, determine the drag factor of the icy highway. Round to the nearest hundredth.
To determine the drag factor of the icy highway, we can use the centrifugal force formula:
F = m * ω^2 * r
Where:
F is the centrifugal force
m is the mass of the car
ω (omega) is the angular velocity (in radians per second)
r is the radius of the yaw mark
First, we need to convert the speed of the car from km/h to m/s:
80 km/h = (80 * 1000) / 3600 m/s = 22.22 m/s
Next, we need to calculate the angular velocity (ω) using the formula:
ω = v / r
where v is the linear velocity and r is the radius of the yaw mark:
ω = 22.22 m/s / 270 m = 0.0823 rad/s
Now, we can calculate the centrifugal force using the formula:
F = m * ω^2 * r
The force due to gravity can be neglected since we are only considering the centrifugal force:
F = m * ω^2 * r = m * (0.0823 rad/s)^2 * 270 m = 0.729m
The drag factor (f) is defined as the centrifugal force divided by the weight of the car:
f = F / mg
where g is the acceleration due to gravity.
Since the weight (mg) is equal to the mass of the car multiplied by the acceleration due to gravity, we can rewrite the equation as:
f = F / (mg/mg) = F / mg
Assuming the mass of the car (m) to be 1000 kg, we can calculate the drag factor:
f = 0.729m / (1000 kg * 9.8 m/s^2) ≈ 0.0075
Therefore, the drag factor of the icy highway is approximately 0.0075 (rounded to the nearest hundredth).
s = sqrt127fr
where,
s=the speed (km/h) the vehicle was traveling when the brakes were applied
f=the drag factor of the surface
r=radius (m) 0f the curved Yaw mark
a car is traveling at a speed pf 80 km/h on a icy highway. the driver swerves while applying the brakes to avoid a collision. if the Yaw mark that car leaves on the icy asphalt has a radius of 270m, determine the drag factor of the icy highway. Round to the nearest hundredth.
To determine the drag factor (f) of the icy highway, we can rearrange the formula given:
s = sqrt(127 * f * r)
We are given that the speed (s) of the car on the icy highway is 80 km/h, or 80,000 meters per hour. We are also given the radius (r) of the yaw mark left by the car as 270 meters.
Let's substitute these values into the equation:
80,000 = sqrt(127 * f * 270)
To solve for the drag factor (f), we need to isolate it. First, square both sides of the equation:
(80,000)^2 = 127 * f * 270
Simplify the equation:
6,400,000,000 = 34,290 * f
Now, divide both sides of the equation by 34,290 to solve for f:
f = 6,400,000,000 / 34,290 ≈ 186.74
Therefore, the drag factor of the icy highway is approximately 186.74 (rounded to the nearest hundredth).