Answer:
The expression that represents the distance in miles Oliver will travel for h hours is: d=45\times h.
Step-by-step explanation:
The first thing to take into account is the speed equation. It is , where v is the speed, x is the distance and t is the time.
Now, let's change the name of the variables to adapt the equation to the problem. The equation could be converted to:
, where v is the speed in miles per hour, d is the distance in miles and h is the time in hours.
The next step is to manipulate a little bit the previous equation to find the distance in terms of speed (v) and time (h). If you pass the variable time (h) to multiply to the left side with speed (v), the equation would be:
The previous equation is the distance in miles in terms of speed in miles per hour and time in hours. Now, you must replace the value of the speed that is 45 miles per hour. The expression would be:
As the speed is in miles per hour and the time is in hours, the distance expression is in miles. So if you replace any value for h (in hours), you will obtain the distance in miles.
Thus, the expression that represents the distance in miles Oliver will travel for h hours is: d = 45 x h.
Step-by-step explanation:
Order of congruence does not matter. For any angles A,B, and C , if ∠A≅∠B and ∠B≅∠C , then ∠A≅∠C . If two angles are both congruent to a third angle, then the first two angles are also congruent.
2 (1/3) + 3 (1/3) is easy to add
2 + 3 = 5
(1/3) + (1/3) = (2/3)
So, it equals 5 (2/3)
She needs 10 - 5 (2/3) more
= 4 (1/3)
secx=1/cosx
1. To solve this problem you must apply the following proccedure:
2. You have to verify if secx sinx=tanx, so, you have:
3. You know that secx=1/cosx, and tanx=sinx/cosx, therefore:
(secx)(sinx)
(1/cosx)(sinx)
sinx/cosx
tanx
9514 1404 393
Answer:
9.5°, yes
Step-by-step explanation:
The relevant trig relation is ...
Tan = Opposite/Adjacent
The distance opposite the angle of elevation is the plane's height, 500 m. The distance adjacent to the angle of elevation is the horizontal distance to the plane, 3 km = 3000 m. Then the angle is found from ...
tan(α) = 500/3000 = 1/6
α = arctan(1/6) ≈ 9.46°
The plane is approaching at an angle of 9.46°. It is safe to land, since that angle is less than 15°.
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<em>Additional comment</em>
The usual descent angle for most commercial air traffic is 3°. Some airport geography demands it be different (steeper). A higher descent angle can put undue stress on the landing gear.
