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3 years ago

0.03256 3 s.f help please

Mathematics

1 answer:

Andrews [41]3 years ago

4 0

Answer:

Step-by-step explanation:

This is hard

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What is (40x33)+(4)x4??<br> Good luck

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PLZ let me know if im wrong!!!

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3 years ago

Whats the answer and steps to 5g+4(-5+3g)=1-g

Lady_Fox [76]

$5g+4(-5+3g)=1-g \\\\ 5g-20+12g=1-g \\\\ 17g-20=1-g\\\\ 17g+g=1+20\\\\18g=21 \\\\ \boxed{g=\frac{21}{18}=\frac{7}{6}}$

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3 years ago

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Evaluate the trigonometric ratio

vfiekz [6]

Answer:

Step-by-step explanation:

Given $\lim_{x \to 0} \frac{sinxcosx}{x}$

multiply the numerator and denominator with 2

we get $\lim_{x \to 0} \frac{2sinxcosx}{2x}$

We know that sin2x=2sinxcosx

$\lim_{x \to0} \frac{sin2x}{2x}$

Now consider y=2x

As x→0 2x→0 y→0

Now limit changes to

$\lim_{y \to0} \frac{siny}{y}$ =1

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3 years ago

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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4 years ago

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Butoxors [25]

Perimeter=20 and the area=25

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4 years ago

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