^ 3 sqrt 750 + ^ 3 sqrt 2058 - ^ 3 sqrt 48
Rewriting the expression we have
^ 3 sqrt (6 * x ^ 3) + ^ 3 sqrt (6 * y ^ 3) - ^ 3 sqrt (6 * z ^ 3)
That is, we have the following equations:
6 * x ^ 3 = 750
6 * y ^ 3 = 2058
6 * z ^ 3 = 48
Clearing x, y and z we have:
x = 5
y = 7
z = 2
Then, rewriting the expression
x (^ 3 sqrt (6)) + y (^ 3 sqrt (6)) - z (^ 3 sqrt (6))
Substituting the values
5 (^ 3 sqrt (6)) + 7 (^ 3 sqrt (6 *)) - 2 (^ 3 sqrt (6))
10 (^ 3 sqrt (6))
answer
the simple form of the expression is
D) 10 ^ 3 sqrt 6
Length of x is 98.2 m
<u>Step-by-step explanation:</u>
Step 1:
Use the trigonometric ratio tan 27° to find the common side of both the right angled triangles.
tan 27° = opposite side/adjacent side = opposite side/9
∴ Opposite side = 9 tan 27° = 9 × - 3.27 = -29.46 m
Step 2:
Use this side and trigonometric ratio cosine to find the value of x.
cos 49° = adjacent side/x = -29.46/x
∴ x = -29.46/cos 49° = -29.46/0.30
= 98.2 m (negative value neglected)
Answer:
48 mph
Step-by-step explanation:
First we need to find the distance from Elkhart to Chicago. Toledo to Elkhart is 136 miles and Toledo to Chicago 244 miles.
So the distance from Elkhart to Chicago can be calculated, since Chicago is farther from Toledo than Elkhart, as: distance(Toledo to Chicago) - distance(Toledo to Elkhart). These distances are given in the problem, so the distance from Elkhat to Chicago is: 244 miles - 136 miles = 108 miles.
This problem basically wants to know the slowest you can be yet still ariving on time. If you are the minimum speed, you will arrive in Chicago exactly at 10:30 A.M. So you have 2 hours and 15 minutes(10:30 A.M - 8.15A.M.) to drive 108 miles.
15 minutes is a fourth of a hour, so you have 2.25hours to go through 108 miles.
The minimum speed you must maintain is 108mph/2.25h = 48mph.
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