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Olin [163]
3 years ago
5

Lavonne sold 4 times as many raffle tickets as kenneth. lavonne sold 56 raffle tickets. how many tickets did kenneth sell?

Mathematics
2 answers:
bonufazy [111]3 years ago
7 0

56÷4=14

Kenneth sold 14 tickets 
torisob [31]3 years ago
5 0

14 is the answer. Just divide 56/4 and you get 14.

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Acceleration  =  (change in speed) / (time for the change) .

Change in speed = (25 m/s - 9 m/s) =  16 m/s

Time for the change = (8 sec  -  5 sec)  =  3 seconds

Acceleration  =  (change in speed) / (time for the change)

                       =                 (16 m/s)  /  (3 sec) 

                       =                          (5 and 1/3)  m/s²  .

(rounded to the nearest hundredth ...  5.33 m/s² . )
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Answer:

x = 2

Step-by-step explanation:

-8(x-6) = 32 (distribute)

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The correct answer is B) 1.15 seconds.
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A particle is projected with a velocity of <img src="https://tex.z-dn.net/?f=40ms%5E-%5E1" id="TexFormula1" title="40ms^-^1" alt
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Answer:

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Step-by-step explanation:

The vertical component of the initial launch can be found using basic trigonometry. In any right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse. Let the vertical component at launch be y. The magnitude of 40\text{ m/s} will be the hypotenuse, and the relevant angle is the angle to the horizontal at launch. Since we're given that the angle of elevation is 60^{\circ}, we have:

\sin 60^{\circ}=\frac{y}{40},\\y=40\sin 60^{\circ},\\y=20\sqrt{3}(Recall that \sin 60^{\circ}=\frac{\sqrt{3}}{2})

Now that we've found the vertical component of the velocity and launch, we can use kinematics equation v_f^2=v_i^2+2a\Delta y to solve this problem, where v_f/v_i is final and initial velocity, respectively, a is acceleration, and \Delta y is distance travelled. The only acceleration is acceleration due to gravity, which is approximately 9.8\:\mathrm{m/s^2}. However, since the projectile is moving up and gravity is pulling down, acceleration should be negative, represent the direction of the acceleration.

What we know:

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Solving for v_f:

v_f^2=(20\sqrt{3})^2+2(-9.8)(50),\\v_f^2=1200-980,\\v_f^2=220,\\v_f=\sqrt{220}=\boxed{2\sqrt{55}\text{ m/s}}

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