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2 years ago

^PLZ HELP, I NEED IT BY TODAY!!! (PART B and PART C only) PLZ THANKS

Mathematics

1 answer:

Lana71 [14]2 years ago

3 0

Part B (Taxi):

12 miles = 0.5 miles + (115×0.1) miles

Rate: $2.40 + (115×0.20)

= $25.40

Total cost (with tip): $25.40 + $2

= $27.40

20 miles = 0.5 miles + (195×0.1) miles

Rate: $2.40 + (195×0.20)

= $41.40

Total cost (with tip): $41.40 + $2

= $43.40

50 miles = 0.5 miles + (495×0.1) miles

Rate: $2.40 + (495×0.20)

= $101.40

Total cost (with tip): $101.40 + $2

= $103.40

Part B (Uber):

12 miles (rate) = $1 + ($1.05×12)

= $13.60

Total cost (with tip): $13.60 + $2

= $15.60

20 miles (rate) = $1 + ($1.05×20)

= $22

Total cost (with tip): $22 + $2

= $24

50 miles (rate) = $1 + ($1.05×50)

= $53.50

Total cost (with tip): $53.50 + $2

= $55.50

Part C

Uber is a better mode of transportation. This is because given the same distance travelled for both the taxi and the Uber, Uber has a lower cost compared to the taxi.

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A manufacturing plant earned \$80$80dollar sign, 80 per man-hour of labor when it opened. Each year, the plant earns an addition

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$A(t)=80(1.05)^t$

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3 years ago

A projectile is fired with muzzle speed 250 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe

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The projectile's horizontal and vertical positions at time $t$ are given by

$x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ\,t$

$y=30\,\mathrm m+\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ\,t-\dfrac g2t^2$

where $g=9.8\dfrac{\rm m}{\mathrm s^2}$ . Solve $y=0$ for the time $t$ it takes for the projectile to reach the ground:

$30+\dfrac{250}{\sqrt2}t-4.9t^2=0\implies t\approx36.2458\,\mathrm s$

In this time, the projectile will have traveled horizontally a distance of

$x=\dfrac{250\frac{\rm m}{\rm s}}{\sqrt2}(36.2458\,\mathrm s)\approx6400\,\mathrm m$

The projectile's horizontal and vertical velocities are given by

$v_x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ$

$v_y=\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ-gt$

At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about $\sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}$ .

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3 years ago