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Zielflug [23.3K]
3 years ago
11

A golf ball í hit off the ground at an angel pi/6 travels 400ft.how long in the air

Mathematics
1 answer:
steposvetlana [31]3 years ago
7 0
You can write the equations of motion as
   h(t) = -16t² + v₀sin(θ)t
   d(t) = v₀cos(θ)t

Solving the second of these for t and substituting into the first equation gives the time of flight as
   0 = -16t² + d(t)·tan(θ)
   t = √(d(t)·tan(θ)/16)

So, for d(t) = 400 and θ = π/6, this becomes
   t = √((400/16)/√3)
   t = 5/√(√3) ≈ 3.79918

The ball is in the air about 3.80 seconds.
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In a race, the mean time for three runners was 12.4 seconds and the mean
Ne4ueva [31]

Answer:

11.8

Step-by-step explanation:

3(12.4) + 6(11.5) =  37.2 + 69 = 106.2

106.2/9 = 11.8

4 0
3 years ago
What is the function form for 2x + 3y = 12
valina [46]
2x+3y=12
Subtract by 2x
3y=-2x+12
Divide by 3
y=-2/3x+4
4 0
3 years ago
Read 2 more answers
Five cupcakes and two cookies cost $19.75. Two cupcakes and four cookies cost $ 17.50.
alina1380 [7]

Answer:

Step-by-step explanation:

Let x and y represent the cost of a cupcake and cookie respectively.

Given that;

Five cupcakes and two cookies cost  $19.75.

5x+2y=19.75 ---------- 1

Two cupcakes  and four cookies cost $17.50.

2x+4y=17.50 --------------2

Let's solve the simultaneous equation by elimination;

Multiply equation 1 by 2;

10x+4y=39.50-------3\\

Subtract equation 2 from equation 3;

10x-2x+4y-4y=39.50-8x=22

8x=22

divide both sides by 8

\frac{8x}{8} =\frac{22}{8} \\x=2.75

Since we have the value of x, let substitute into equation 1 to get y;

5x+2y=19.75\\5(2.75)+2y=19.75\\13.75+2y=19.75\\2y=19.75-13.75\\2y=6\\y=\frac{6}{2} \\y=3.00

therefore , the cost of cupcakes and cookies are;

cupcakes=  2.75\\cookies= 3.00

PLEASE MARK ME AS BRAINLIEST

6 0
2 years ago
Anne is taping 7 boxes for storage. She uses feet of tape to close each box. How much tape will Anne need to tape the boxes? Exp
aleksandrvk [35]
D is the correct answer




hope this helps
7 0
3 years ago
Here's another geometry thing I suck at...
solniwko [45]
A = 133; b = 31; c = 82; d = 64.
Opposite angles in an inscribed quadrilateral are supplementary; this means that d + 116 = 180.  Subtracting 116 from both sides, we have d = 64.

By the same theorem, c + 98 = 180; subtracting 98 from both sides, we have c = 82.

Inscribed angles are equal to 1/2 the measure of the intercepted arc.  Using this, we have

116 = 1/2(a+99)

Multiplying both sides by 2, we have 
232 = a+99

Subtract 99 from both sides:
232 - 99 = a + 99 - 99
133 = a

We also have that 
82 = 1/2(133+b)

Multiplying both sides by 2, we have:
164 = 133 + b

Subtract 133 from both sides:
164 - 133 = 133 + b - 133
31 = b
6 0
3 years ago
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