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3 years ago

11

I have a shape. one side is 13cm, the opposite side is 9cm, the top is 11cm, and the bottom is 10 cm. is it a parallelogram or a

trapezoid

1 answer:

Shtirlitz [24]3 years ago

3 0

It would be a trapezoid because parallelograms have the top and bottom usually the same number.

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Yes

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Find the equation of the line passing through(1,2) and making an angle of 30 degree with y intercept

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Equation of the straight line passing through the point ( 1,2 ) and having the slope

x - √3y -1+2√3 =0

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that the point (1,2) and an angle = ∝ = 30°

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$m = \frac{1}{\sqrt{3} } }$

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Equation of the straight line passing through the point ( 1,2 ) and having the slope

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Given s left parenthesis t right parenthesis equals 5 t squared plus 5 ts(t)=5t2+5t, find

finlep [7]

Answer:

The velocity function is $v(t)=10t+5$ .

The acceleration function is $a(t)=10$ .

When t = 44, the velocity is $v(44)=445 \:\frac{ft}{s}$ .

When t = 44, the acceleration is $a(44)=10\: \frac{ft}{s^2}$ .

Step-by-step explanation:

We know that the position function is given by

$s(t)=5t^2+5t$

Velocity is defined as the rate of change of position or the rate of displacement. If you take the derivative of the position function you get the instantaneous velocity function.

$v(t)=\frac{ds}{dt}$

Acceleration is defined as the rate of change of velocity. If you take the derivative of the instantaneous velocity function you get the instantaneous acceleration function.

$a(t)=\frac{dv}{dt}$

The instantaneous velocity function is given by

$v(t)=\frac{d}{dt} s(t)=\frac{d}{dt}(5t^2+5t)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(5t^2\right)+\frac{d}{dt}\left(5t\right)\\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}\\\\v(t)=10t+\frac{d}{dt}\left(5t\right)\\\\\mathrm{Apply\:the\:common\:derivative}:\quad \frac{d}{dt}\left(t\right)=1\\\\v(t)=10t+5$

The instantaneous acceleration function is given by

$a(t)=\frac{dv}{dt} =\frac{d}{dt}(10t+5)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\a(t)=\frac{d}{dt}\left(10t\right)+\frac{d}{dt}\left(5\right)\\\\a(t)=10$

To find the velocity and acceleration when t = 44, we substitute this value into the velocity and acceleration functions

$v(44)=10(44)+5\\v(44)=445 \:\frac{ft}{s}$

$a(44)=10\: \frac{ft}{s^2}$

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Answer:

The answer is 45 degrees

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