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3 years ago

Help with formulas and explain

Mathematics

1 answer:

Vlad [161]3 years ago

6 0

A. T. S.A =2* AREA OF TRIANGLE +PREIMETER*5.56

VOLUME=AREA OF *LENGTH[ 5.56]

B. T. S. A=πR{R + SLANT HIGHT }

VOLUME=1/3*π*R*HIGHT{ 7 }

volume of a sphere is 3 by 4 by R cube and surface area of a sphere is 4 Pi R Square

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2 years ago

Two drivers, Alison and Kevin, are participating in a drag race. Beginning from a standing start, they each proceed with a const

Vilka [71]

Answer:

Alison wins against Kevin by 0.93 s

Step-by-step explanation:

Alison covers the last 1/4 of the distance in 3 seconds, at a constant acceleration $a_a$ , we have the following equation of motion

$s/4 = a_at_a^2/2$

where s (m) is the total distance, ta = 3 s is the time

$s = 4a_a3^2/2 = 18a_a$

$a_a = s/18$

Similarly, Kevin overs the last 1/3 of the distance in 4 seconds, at a constant acceleration $a_k$ , we have the following equation of motion:

$s/3 = a_kt_k^2/2$

tk = 4 s is the time

$s = 3a_k4^2/2 = 24a_k$

$a_k = s/24$

Since $a_a = s/18$ we can conclude that $a_a > a_k$ , so Alison would win.

The time it takes for Alison to cover the entire track

$s = a_aT_a^2/2$

$T_a^2 = 2s/a_a = 2s/(s/18) = 36$

$T_a = \sqrt{36} = 6 s$

The time it takes for Kevin to cover the entire track

$s = a_kT_k^2/2$

$T_k^2 = 2s/a_k = 2s/(s/24) = 48$

$T_a = \sqrt{48} = 6.93 s$

So Alison wins against Kevin by 6.93 - 6 = 0.93 s

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2 years ago

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mina [271]

let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill$

$\bf cos(\theta )=\cfrac{\stackrel{adjacent}{\pm\sqrt{11}}}{\stackrel{hypotenuse}{6}} \\\\\\ tan(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{adjacent}{\pm\sqrt{11}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{tan(\theta )=\pm\cfrac{5}{\sqrt{11}}\cdot \cfrac{\sqrt{11}}{\sqrt{11}}\implies tan(\theta )=\pm\cfrac{5\sqrt{11}}{11}}$

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2 years ago