An item priced at $10.20 has a sales tax of $0.41.Find the sales tax rate expressed as a percent.

The height h (in feet) of an object dropped from a ledge after x seconds can be modeled by h(x)=−16x2+36 . The object is dropped

kakasveta [241]

Check the picture below.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}$

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

$\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}$

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2 = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill$

quick info:

in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².

4 0

3 years ago

at the metal-casting company where i work, i must set up a conveyor system to move castings from the furnace to the heat treatme

Harman [31]

Maximum time allowed = 7 deg. / 0.03 deg/s = 233.3 s
Minimum speed = 55 yd * 3 ft/yd / 233.3 s
= 0.707 ft/s

4 0

4 years ago

Solving Exponential Equations (lacking a common base)<br><br>(0.52)^q=4

bulgar [2K]

Answer:

q = log4 / log0.53

q = -2.189

Step-by-step explanation:

(0.53)^q = 4

Taking log of both sides!

q log 0.53 = log 4

q = log 4 / log 0.53

q = 0.602 / -0.275

q = - 2.189

This can be checked to confirm correctness.

Substituting q = -2.189

(0.53)^ -2.189 = 1/ 0.249

= 4.01 Proved!

(Note that the "1" is because of the negative sign)

8 0

3 years ago

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Pls help this is a grade?!?!?!?

maksim [4K]

Net 2....................

5 0

3 years ago

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A tourist boat is used for sightseeing in a nearby river. The boat travels 2.4 miles downstream and in the same amount of time,

topjm [15]

Answer: the current of the river is 3 mph.

Step-by-step explanation:

Let x represent the current of the river.

The boat travels 2.4 miles downstream. Assuming it travelled with the current of the river. If the boat travels at an average speed of 21 miles per hour in the still water, it means that the total speed at which the boat travelled downstream is

(21 + x) mph

Time = distance/speed

Time taken to travel downstream is

2.4/(21 + x)

it travels 1.8 miles upstream. Assuming it travelled against the current of the river. it means that the total speed at which the boat travelled upstream is

(21 - x) mph

Time taken to travel upstream is

1.8/(21 - x)

Since the time taken to travel upstream and downstream is the same, then

2.4/(21 + x) = 1.8/(21 - x)

Cross multiplying, it becomes

2.4(21 - x) = 1.8(21 + x)

50.4 - 2.4x = 37.8 + 1.8x

2.4x + 1.8x = 50.4 - 37.8

4.2x = 12.6

x = 12.6/4.2

x = 3 mph

7 0

3 years ago

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