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A = (1/2)bh   ...  so.....

36 + 5/6 =  (1/2)(6 + 1/2) h       and 36 + 5/6 =  221/6   and 6 + 1/2  =  13/2   ...so.....

221/6  = (1/2)(13/2) h

221/6  = (13/4) h     multiply both sides by 4/13

(221/6)(4/13)  = h

(221/13)(4/6)  = h

(221/13)(2/3)  = 442 / 39  = 34/3 ft  = h = (11 + 1/3) ft

Typo. meant 2*(x-3) and 7*(x+2)

Multiply 2 by (x-3) and 7 by (x-2)

We would multiply 2 by both x and -3, like so, getting...

2 * x = 2x and 2*-3 = -6

2x-6

then we multiply 7 by both x and 2

7*x = 7x and 7*2 = 14

This equals 7x +14


Bring those together and we have: 2x -6 + 7x +14

Combine like terms (add the numbers with x's and the number without an x)

2x + 7x = 9x

-6 + 14 = 8

The answer is 9x + 8, as far as I see. 

If 4 assistants take 60 minutes to do the job, then one assistant would take four times as long = 240 min

So 25 assistants could do the job in 1/25 of the time.......so ......  240* 1/25 = 9.6 minutes

Thank you CPhill!

2 - 12/(4 - x)  = (2x + 4) / (x^2 - 16)   we can rewrite this as

2 + 12/(x - 4)   = 2(x + 2)/ [(x - 4) (x + 4) ]        multiply through by [(x - 4) (x + 4) ]

2 [(x - 4) (x + 4) ] + 12(x + 4)  = 2(x + 2)        divide thrrough by 2

[(x - 4) (x + 4)] + 6(x + 4)  = x + 2    simplify

x^2 - 16 + 6x + 24  =  x + 2   

x^2 + 6x + 8   = x + 2

x^2 + 5x + 6    =  0      factor

(x + 3) (x + 2)  = 0      and setting each factor to 0 we have x = -2   or  x = -3 

{There are no extraneous solutions }

3*(x-4.5)⁵ = 80

Divide both sides by 3

(x - 4.5)⁵ = 80/3

Take the fifth root of both sides

x - 4.5 = (80/3)^1/5

Add 4.5 to both sides

$${\mathtt{x}} = {\left({\frac{{\mathtt{80}}}{{\mathtt{3}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{5}}}}\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{4.5}} \Rightarrow {\mathtt{x}} = {\mathtt{6.428\: \!385\: \!008\: \!005\: \!254\: \!5}}$$

.

sin(angle) = opposite/hypotenuse

so opposite = hypotenuse*sin(angle)

$${\mathtt{X}} = {\mathtt{74}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{32}}^\circ\right)} \Rightarrow {\mathtt{X}} = {\mathtt{39.214\: \!025\: \!553\: \!242}}$$

.

Wow.. it was that easy?! Thanks so much Alan!

Use them on lots of practice problems.

Keep the following saying in mind:

I hear and I forget.

I see and I remember.

I do and I understand.

.

Or just:

f(1) = -4

f(2) = f(1) + 3   so f(2) = -4 + 3  or f(2) = -1

. 

Add 7 to both sides:

41 + 7 = 12*d

48 = 12*d

Divide both sides by 12

48/12 = d

or

d = 4

.

Let d stand for number of dogs and c stand for number of cats.

d = 2*c + 12                ...(1)

d + c = 75                   ...(2)

Replace d in (2) using (1)

2*c + 12 + c = 75

3*c + 12 = 75

3*c = 75 - 12

3*c = 63

c = 63/3

c = 21

So there are 21 cats  (and 2*21 + 12 = 54 dogs)

.

Simply replace t by 70:

$${\mathtt{y}} = {\mathtt{40}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}^{\left({\frac{{\mathtt{70}}}{{\mathtt{40}}}}\right)} \Rightarrow {\mathtt{y}} = {\mathtt{11.892\: \!071\: \!150\: \!027\: \!210\: \!7}}$$

y = 11.89 grams to the nearest one-hundredth of a gram

.

Careful Anonymous; you've gone wrong in the third line. 

This should be:

x = 30 + 15

so x = 45

.

it is 1/m^3

P(t) = -2t²+100t+2500

P'(t) = -4t +100

-4t+100 = 0                =>  100=4t            =>           t=25

maximum point ( 25 ; 3750 )

An HD television has a list price of $752. If a series discount of 10/15 is offered, what is the net cost after the series discount? Use the method of complements to solve this problem.

The List price = $752,      The series discount is: 10/15   (the complements of 10/15 are: 0.90 and 0.85)

The Net cost equivalent is calculated this way:     0.90 * 0.85 = 0.765                                     

Net cost = List price x Net cost equivalent

Net cost = 752 * 0.765 

              = $ 575.28

$${\mathtt{V}} = {\frac{{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{3.14}}{\mathtt{\,\times\,}}{{\mathtt{r}}}^{{\mathtt{3}}}}{{\mathtt{3}}}}$$

$${{\mathtt{r}}}^{{\mathtt{3}}} = {\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{V}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{3.14}}\right)}}$$             =>       $${\mathtt{r}} = {\sqrt[{{\mathtt{{\mathtt{3}}}}}]{{\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{V}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{3.14}}\right)}}}}$$

$${\sqrt[{{\mathtt{{\mathtt{3}}}}}]{{\frac{{\mathtt{3}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{3.14}}\right)}}}} = {\mathtt{0.620\: \!455\: \!356\: \!763\: \!528}}$$   

               $${\mathtt{r}} = {\mathtt{0.620\: \!455}}{\mathtt{\,\times\,}}{\sqrt[{{\mathtt{{\mathtt{3}}}}}]{{\mathtt{V}}}}$$

4*(18-x) = 20x                   | : 4

      18-x = 5x                       |  + x

   18 = 6 x               =>     x =3

hi,

this problem is a simple substitution.

so you know that y = 15, so substitute it into the equation you were given!

$${\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{y}} = {\mathtt{30}}$$

$${\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{15}} = {\mathtt{30}}$$

so now with 15 substituted in, all you have to do is move the 15 over to the other side!

$${\mathtt{x}} = {\mathtt{30}}{\mathtt{\,-\,}}{\mathtt{15}}$$

and now simplify by subtracting 15 from 30!

$${\mathtt{x}} = {\mathtt{15}}$$

hope this helps!

using n notation is just a fancy way of describing a sequence.

n is the number you're currently at, so with that logic n+1 is the term after it in the sequence.

f(n) is the function using n as the input.

so you know that the first term of the sequence is -4, or f(n)= -4.

hi,

1 gigabyte is 1000 megabytes, so all you have to do in order to find your storage amount in megabytes is multiply your gigabyte by 1000.

$${\mathtt{1.9}}{\mathtt{\,\times\,}}{\mathtt{1\,000}} = {\mathtt{1\,900}}$$

and so 1.9 GB times 1000 MB in 1 GB is 1900 MB!

hope this helps!

$${\sqrt{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}} = {\mathtt{3.414\: \!213\: \!562\: \!373\: \!095}}$$

what is the mean of a set of numbers

Find the mean of the set A = {3,-2,7,1,-5,8}

                            $${\mathtt{mean}} = {\frac{\left({\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}}\right)}{{\mathtt{6}}}}$$

                            $${\mathtt{mean}} = {\mathtt{2}}$$