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3 years ago

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Which number rounds to 65.8 when rounded to the nearest tenth

1 answer:

e-lub [12.9K]3 years ago

5 0

Answer:

65.8

Step-by-step explanation:

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An object is launched at 28 meters per second (m/s) from a 336-meter tall platform. The equation for the object's height s at ti

kondaur [170]

Answer:343 m

Step-by-step explanation:

Given

launch velocity of object is $u=28\ m/s$

height of Platform $h=336\ m$

height of object is given by

$s(t)=-7t^2+14t+336$

For maximum height velocity of object is zero

i.e. $\frac{ds}{dt}=0$

$\frac{ds}{dt}=-14t+14+0=0$

$t=\frac{14}{14}=1\ s$

Therefore after 1 sec object achieves maximum height

$s(1)=-7(1)+14(1)+336$

$s(1)=343\ m$

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4 years ago

PLEASE HELP ME FIND X please I really need help

Liono4ka [1.6K]

Answer:

is this a multiple-choice type thing or type-in? If multiple-choice, what answers are given?

Step-by-step explanation:

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3 years ago

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What is 3 4/6 - 2 3/6

tino4ka555 [31]

The answer is 1 1/6.
Hope it was helpful :)

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3 years ago

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For what values of x is the expression below defined??<br> Square root of 2-x

PIT_PIT [208]

I don't know, Sorry, but I will keep looking at it to see if I can figure it out, Sorry

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3 years ago

A planet follows an elliptical path described by 16 x squared + 4 y squared = 64. A comet follows the parabolic path y equals x

saul85 [17]

Answer:

The comet might intersect the orbiting planet at (0, -4), $(\pm\frac{\sqrt{31}}{2},\frac{15}{4})$ .

Step-by-step explanation:

Given that, a planet follows an elliptical path described by

$16x^2+4y^2=64$ .........(1)

A comet follows the parabolic path

$y=x^2-4$

$x^2=y+4$ .........(2)

To find the intersecting point, we need to solve the above equations.

Putting $x^2=y+4$ in the equation (1)

$y+4+4y^2=64$

$\Rightarrow 4y^2+y+4-64=0$

$\Rightarrow 4y^2+y-60=0$

$\Rightarrow 4y^2+16y-15y-60=0$

$\Rightarrow 4y(y+4)-15(y+4)=0$

$\Rightarrow(y+4)(4y-15)=0$

$\Rightarrow y=-4, \frac{15}{4}$

When y = -4 , then $x^2=-4+4$ $\Rightarrow x=0$

When $y=\frac{15}{4}$ , then $x^2=\frac{15}{4}+4$ $\Rightarrow x=\pm \frac{\sqrt{31}}{2}$

The comet might intersect the orbiting planet at (0, -4), $(\pm\frac{\sqrt{31}}{2},\frac{15}{4})$ .

5 0

3 years ago