Answer: 5/36
Step-by-step explanation:
That is the answer
Answer:
a) Brain sweep out 4 radians
b) Brian is -0.3922 km to the right of the center of the ski trail
c) Brian is -0.4541 km above of the center of the ski trail
Step-by-step explanation:
Given the data in the question;
circular ski trail radius = 0.6 km
Derek starts at the 3-o'clock position and travels 2.4 km in the counter-clockwise direction.
a) How many radians does Brian sweep out?
arc length = 2.4 km
from the diagram;
arc length = rθ
θ = arc length / r
θ = 2.4 / 0.6
θ = 4 radians
Therefore, Brain sweep out 4 radians
b) When Brian stops skiing, how many km is Brian to the right of the center of the ski trail?
from the diagram;
x = rcosθ
we substitute
x = 0.6 × cos(4)
x = 0.6 × -0.65364
x = -0.3922 km
Therefore Brian is -0.3922 km to the right of the center of the ski trail
c) When Brian stops skiing, how many km is Brian above of the center of the ski trail?
from the diagram;
m = rsinθ
m = 0.6 × -0.7568
m = -0.4541 km
Therefore, Brian is -0.4541 km above of the center of the ski trail
For every 140 miles, Trey uses 4 gallons:
140:4
We can simplify this ratio to 1 gallon by dividing each side by 4
35 miles : 1 gallon
1) To find out the miles travelled in 6 gallons, we multiply ea b side by 6
35*6 = 210miles : 1*6 = 6gallons
So it is false.
2) For 5 gallons, we multiply each side by 5
35*5 = 175miles : 1*5 = 5gallons
Once again, it is false
3) We modify the ratio for price
35 miles : 1 gallon
35 miles : 3.45 dollars
And then we work in reverse - instead of multiplying the ratio to reach a certain number, we divide a certain number by the ratio
$27.6/3.45 = 8
For the question to be true, each side has to become
260 miles : 27.6 dollars when multiplied by 8
However, 35*8 = 280, not 260, so the statement is false.
Answer:
(x < 0) ∪ (x > 1/2)
Step-by-step explanation:
<u>First inequality</u>:
x^3 < 0
x < 0 . . . . . take the cube root
<u>Second inequality</u>:
4x^4 > x^2
4x^4 -x^2 > 0
x^2(4x^2 -1) > 0
Any non-zero x will make the first factor positive. The product will be positive only when the second factor is positive, for x^2 > 1/4
x^2 > 1/4
|x| > 1/2 . . . . . take the square root
x < -1/2 or x > 1/2
The first part of this solution is already covered by the solution to the first inequality. Hence the solution set for the system of inequalities is ...
(x < 0) ∪ (x > 1/2)
