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Find the percent of markup. Round to the nearest tenth. $33.00 to $45.00

$33  +  (percent of markup * $33)   =   $45         Subtract  $33  from both sides of this equation.

percent of markup * $33   =   $12       Divide both sides by  $33 .

percent of markup   ≈   0.364   =   36.4/100   =   36.4%

Estimate an 18% tip for a 63.50 meal. 

It says to estimate...so let's say that is about  20% of 60

20%  of  60   =   20/100  of  60   =   2/10  of  60   =   2  *  60/10   =   2 * 6   =   12

A new law requires that 12% of an individual's income be invested in the stock market. Your accounts show that you need to put $420 in the stock market this year. How much did you earn this year.

12%  of your income  =  amount you need to put in the stock market

12%  of your income  =  $420

12/100 of your income  =  $420

12/100  *  your income  =  $420

0.12  *  your income   =  $420      Divide both sides of this equation by  0.12  .

your income   =   $420 / 0.12

your income   =   $3500

First let's think about a simpler problem:     Find  3/4  of  20  .

We want to take  20  , divide it into  4  parts of equal amount, and take  3  of them.

20/4  =  5   ( Think...  5 + 5 + 5 + 5  =  20 )

So... one fourth of  20  is  5  .  Three fourths of  20  is  3 * 5 , or  15 .

3/4  of  20   =   3 * 20/4   =   3 * 5   =   15

Using the same logic.... we can find  21/25  of  $27,000 .

21/25  of  $27,000   =   21 *  $27,000/25   =   21  *  $1080   =   $22,680

After  t  seconds, the height of the sphere above the ground is  h  feet, where

h  =  -16t² + 30t + 78

After how many seconds does the sphere hit the ground?

When the sphere is on the ground, its height is zero. We want to know the time when the height of the sphere is zero. That is, we want to know  t  when  h = 0 . We want to know the value of  t  that makes this equation true:

0  =  -16t² + 30t + 78

                                          To solve for  t , let's first divide both sides of the equation by  2 .

0  =  -8t² + 15t + 39

                                          Now let's solve this quadratic equation using the quadratic formula.

t  =  \({-15 \pm \sqrt{15^2-4(-8)(39)} \over 2(-8)}\)

t   =   \({-15 \pm \sqrt{225+1248} \over -16}\)

t   =   \({-15 \pm \sqrt{1473} \over -16}\)

t   =   \({-15 + \sqrt{1473} \over -16}\)          or          t  =  \({-15 - \sqrt{1473} \over -16}\)

t  ≈  -1.461                    or          t  ≈  3.336

Since  t  is a measure of time, it should be the positive option.

So...the sphere hits the ground after about  3.336  seconds.

a)  For the last pair, here's an example and explanation.

Imagine someone is putting air in a spherical balloon. The radius of the balloon is increasing as time passes. Let's say we have a function that can tell us the radius at any point in time....

the radius of the balloon  =  g(x)   , where  x  is the time in some units, let's say seconds.

So if someone asked, "What is the radius of the balloon when the time is  7  seconds?"

It would just be  g(7) .

We also have a function that tells us the volume, if we know what the radius is....

the volume of the balloon  =  h(x)  , where  x  is its radius in inches.

So if someone asked, "What is the volume of the balloon when the radius is  3 inches?"

It would just be  h(3) .

Now what if someone asked, "What is the volume of the balloon when the time is  7  seconds?"

We know the radius at any given time:  When the time is  7 , the radius is  g(7) .

We know the volume at any given radius:  When the radius is  g(7) , the volume is  h( g(7) ).

So when the time is  7  seconds, the volume is  h( g(7) )

So....

h( g(x) )  =  the volume as a function of time.

Question 2

a)    3 - 3^x   =   3 - f(x)   =   g( f(x) )   =  g o f

b)    x² - 6x + 6   =   x² - 6x + 9 - 3   =   (3 - x)² - 3   =   ( g(x) )² - 3   =   h( g(x) )   =   h o g

c)    6 - x²   =   3 + 3 - x²   =   3 - x² + 3   =   3 - (x² - 3)   =   3 - ( h(x) )   =   g( h(x) )   =   g o h

Question 1

a)     f( x )  =  3x² - 5x + 6    and    g(x)  =  x² + 3x

f o g   =   f( g(x) )

f o g   =   f( x² + 3x )

f o g   =    3( x² + 3x )² - 5( x² + 3x ) + 6    I'll let you simplify these if necessary...

Thanks!! Glad to be of some help!!

num of cats  =  n

num of fleas per cat  =  2n

total num of fleas  =  (num of fleas per cat) * (num of cats)   =   2n * n   =   2n²

If we take the total number of fleas, subtract the fleas of two cats, then subtract  3  for every remaining cat, we get half the total number of fleas.

(total num of fleas) - 2(num of fleas per cat) - 3(n - 2)  =  (total num of fleas)/2

2n²  -  2(2n)  -  3(n - 2)   =   (2n²)/2

2n² - 4n - 3n + 6   =   n²

                                                 Subtract  n²  from both sides and combine like terms.

n² - 7n + 6  =  0

                                                 Factor the left side.

(n - 6)(n - 1)  =  0

n = 6      or      n = 1

n  must be  6  because we can't remove  2  cats if there is only  1  cat to begin with.