Answer:
566 ft
Step-by-step explanation:
Given the polynomial equation
-16t² + 300t + 30 = 0
We're asked to find how high the rocket would rise in a 2 second fuse rise. This is a simple instruction that can be obtained by setting the unknown, t, in the equation to 2. Doing this, we have
-16 * 2² + 300 * 2 + 30
-16 * 4 + 600 + 30
-64 + 630 = 566 ft
Then, we can therefore conclude that the rocket will rise high up to a distance of 566 ft
The distance between the points A to B is 899.9 feet. After rounding off the nearest integer we get 900 feet as the final answer.
Given we know that CD is perpendicular to AD.
The distance between CD is 139 feet.
As from points A the boat's crew measure the angle of elevation to the beacon as 6°
therefore, m∠A = 6°
Another time the angle of elevation is measured from point B which is 19°.
therefore, m∠DBC = 19°
tan 19° = CD/BD
BD = CD/tan19°
BD = 136/tan 19°
now for tan 6° = CD/AD (tangent is opposite over adjacent)
AD = CD/tan 6°
AD = 136/tan 6°
AB = AD ₋ BD
AB = 136/tan 6° ₋ 136/tan 19°
AB = 1295.2 ₋ 395.3
AB = 900 feet
hence the distance from point A to B is 900 feet.
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7 times 4 =28 this is cuz there is seven days in a week and 7x4
5x+15-6+7x
15-6= 9.
5=7= 12
12x+9 .
The solution to the statement 4 + (-4) is correctly given by the third number line.
- The solution given by plot 1 is : (-4) + (-4) as both arrows points 4 units to the left.
- The solution given by plot 2 is : ( 4 + 4) as both arrows points 4 units to the right.
- The solution given by plot 3 is (4) + (-4) as one arrow points 4 units to the left and the other points 4 units to the right.
- The solution given by the plot 4 is (-4 + 8) as one arrow points 4 units to the left and the other points 8 units to the right.
Therefore, the Number line which shows the solution 4 + (-4) is the third number line.
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