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2 years ago

Which of the following methods of timekeeping is the least precise?

Mathematics

2 answers:

katen-ka-za [31]2 years ago

8 0

I think C would be your answer

Verizon [17]2 years ago

5 0

Answer:

na D cause imagine always observing the seasons bro

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What are the co-ordinates meant to be for the graph ... I need answers fast but if you want to be an extra help go on my maths w

Yuki888 [10]

Answer:

Step-by-step explanation:

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3 years ago

What is the length of one side of a cube that has a surface area of 1,734 square mm?

jarptica [38.1K]

Answer:

a = 17

Step-by-step explanation:

1,734 = 6a^2

1,734/6 = 6a^2/6

$\sqrt{289}$ = $\sqrt{a^2}$

a = 17

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3 years ago

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An elevator moving down passes its neighbor, an elevator moving up. Their speed relative to one another is 8 m/s. What is the ve

ANTONII [103]

LiftA= u m/s ( upwards)
LiftB= -u m/s(downward)

Velocity A relative to B= V lift A- V lift B=8m/s
u-(-u)=8
u=4m/s
Lift A= 4 m/sec
Lift B= -4m/sec
For someone standing on first floor will be stationary W.r.t to the lift.
V lift A relative to Man= V liftA - V man= 4m/s

V lift B relative to Man= V lift B- V Man= -4 m/s

5 0

3 years ago

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Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the

Rama09 [41]

Question:

Britney throws an object straight up into the air with an initial velocity of 27 ft/s from a platform that is 10 ft above the ground. Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the object to hit the ground?

1s 2s 3s 4s

Answer:

It takes 2 seconds for object to hit the ground

Solution:

The given equation is:

$h(t) = -16t^2 + v_0t+h_0$

Initial velocity = 27 feet/sec

$h_0 = 10\ feet$

Therefore,

$h(t) = -16t^2 +27t+10$

At the point the object hits the ground, h(t) = 0

$-16t^2 +27t+10 = 0\\\\16t^2-27t-10=0$

Solve by quadratic formula,

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=16,\:b=-27,\:c=-10:\quad \\\\t=\frac{-\left(-27\right)\pm \sqrt{\left(-27\right)^2-4\cdot \:16\left(-10\right)}}{2\cdot \:16}\\\\t = \frac{27 \pm \sqrt{1369}}{32}\\\\t = \frac{27 \pm 37}{32}\\\\We\ have\ two\ solutions\\\\t = \frac{27+37}{32}\\\\t = \frac{64}{32}\\\\t = 2\\\\And\\\\t = \frac{27-37}{32}\\\\t = -0.3125$