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((sin 82°)(cos 37°)) - ((cos 82°)(sin 37°))=sin(45) =1/sqrt(2) =.7071......etc.

The reason the hypotenuse is so "small", is because your angle of 81 degrees is so "big". Remember that sin(90)=1 and sin(81)=0.9877, which is very close to 1. So, the side opposite it is going to be almost as long as the side opposite the 90-degree angle, or the hypotenuse!.

Thanks for your help, it made me realise where I was messing up. After some changing around of the equation, it made more sense to me. 

Basically, I had to change it to ( (18.1 mg/hour^-1) /2 ) / 0.00177 mg/ml = This will get me the answer per hour (5112 ml/hour) as both of the mg get cancelled out in the equation.

Then I have to do (5112 ml/hour) / 60 = to get my answer in minutes. 

Obviously would be easier to convert 9.05 to mg/minutes but then I would not be able to keep track of the 18.1 and 9.05 mg amounts. 

Thanks for your help nonetheless. It really brought clarity to me and helped me figure out where I was going wrong. (i.e. My answer was to be compared to 125ml/min, not have that number in the equation).

the polynomial x^3+ax^2-bx=6 has factors of (x+1) (x+2) what is a and b

I. factors of \((x+1) \quad \Rightarrow x_1 = -1\)

\(\begin{array}{|rcll|} \hline x^3+ax^2-bx -6 &=& 0 \quad &| \quad x_1 = -1\\ ( -1)^3+a( -1)^2-b(-1) -6 &=& 0 \\ -1+a+b -6 & =& 0 \\ a+b -7 & =& 0 \quad &| \quad +7\\ a+b & =& 7 \\ \mathbf{ b} & \mathbf{=}& \mathbf{7-a} \\ \hline \end{array}\)

 II. factors of (x+2) \((x+2) \quad \Rightarrow x_2 = -2\)

\(\begin{array}{|rcll|} \hline x^3+ax^2-bx -6&=& 0 \quad &| \quad x_2 = -2\\ ( -2)^3+a( -2)^2-b(-2) -6 &=& 0 \\ -8+4a+2b -6 &=& 0 \\ 4a+2b -14 &=& 0 \quad &| \quad +14\\ 4a+2b &=& 14 \quad &| \quad : 2\\ 2a+ b &=& 7 \quad &| \quad \mathbf{ b = 7-a} \\ 2a+ 7-a &=& 7 \\ a+ 7 &=& 7 \quad &| \quad x_2 = -7\\ a &=&0 \\ \mathbf{ a} & \mathbf{=}& \mathbf{0} \\\\ b &=& 7-a \quad &| \quad \mathbf{ a} \mathbf{=} \mathbf{0}\\ b &=& 7 - 0 \\ b &=& 7\\ \mathbf{ b} & \mathbf{=}& \mathbf{7 } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline x^3+ax^2-bx &=& 6 \quad &| \quad \mathbf{ a} \mathbf{=} \mathbf{0} \qquad \mathbf{ b = 7 }\\ x^3+0\cdot x^2- 7\cdot x &=& 6 \\ \mathbf{x^3 - 7\cdot x }& \mathbf{=}& \mathbf{6} \\ \hline \end{array}\)

sin(81)=10/x so if .9876 = 10/x, what is the length of the hypotenuse.

why is it so small?  do i need to move a decimal?

((sin 82°)(cos 37°)) - ((cos 82°)(sin 37°))= Sin(82-37) = sin 45 = 1/sqrt2   =    2/sqrt2

I suppose that we're solving for x here.

2 - 3x/8 = 0

-3x/8 = -2

-3x = -16

x = 16/3

If the problem wasn't to solve for x, then I suppose you meant that you wanted this simplified, in which case I believe the following makes a simpler form:

2 - (3x)/8

16/8 - (3x)/8

(16-3x)/8

(2,2), (5,y); slope=2

\(m=\frac{y2-y1}{x2-x1}\)

\(2=\frac{y-2}{5-2}\)

\(y= 2\times{(5-2)}+2\)

\({\color{blue}y=8}\)        !

Remember the order of operations.

(5²*2)-8

(25*2)-8

50-8

42

slope = (y₁-y₂)/(x₁-x₂)

2 = (2-y)/(2-5)

2 = (2-y)/(-3)

-6 = 2-y

-8 = -y

8 = y

x+8=-x

x+8+x=-x+x

2x+8=0

2x+8-8=-8

2x=-8

2x/2=-8/2

x=-4

x+8=-x

\(x+8=-x\)

\(2x=-8\)

\({\color{blue}x=-4}\)        !

Using a two-column proof:

Statements (reasons)

<1 and <2 are complementary, <3 and <4 are complementary, m<1 = m<3 (given)

m<1 + m<2 = 90, m<3 + m<4 = 90 (definition of complementary angles)

m<3 + m<2 = 90, m<3 + m<4 = 90 (substitution property of equality)

m<2 = 90 - m<3, m<4 = 90 - m<3 (subtraction property of equality)

m<2 = m<4 (transitive property of equality)

This formula asks the question: How long does it take for $1 to grow to $3 @ 10% compounded continuously? The answer, just as has been calculated above, is just under 11 years. So, r here =number of years. "n" is normally used to denote number of years, and "r" to denote the "interest rate", but has been reversed here.

She has 1/6 of her original 126 which is $21

Thanks EP   

Last month an orange juice distributor found that the demand for juice was 500 litres each month when the price was $15 per litre. At that time, the price was going up at a rate of $0.50 per month, but due to increased advertising the demand was also going up, at a rate of 30 litres per month. How fast was the revenue increasing?

Let me see,

Last month the revenue was  500*15       =  $7500

This month the revenue will be  530*15.5 =  $8215

8215-7500 = $715/month

That sounds right to me......

oops.....  Melody accidentally used LOG for the last step instead of the natural ln   ....

10*ln(3) = 10.9861

Take the natural ln of both sides;

ln(3)=ln(e^(.1r) )

1.09861 = .1r      Divide both sides by .1

10.9861= r

3 = e^0.1r, find r

Perhaps you mean   3 = e^(0.1r), find r

\(3 = e^{0.1r}\\ ln3=0.1r\\ r=ln3/0.1\\ r=10ln3\)

10*log(3) = 4.7712125471966244

8y+2=138

-2       -2

8y=136

8y/8    136/8

y= 17

There are an infinite number of paired solutions.

This is the equation of a line.

You can graph it but you cannot solve it.

I just drew it and thought about each senario.

The last three were eliminated and the first one appears to work.

So ... drum roll .....

The answer is A

I = (2e^-.10x1) + (2e^-.10x2) = 3.447

F₀₌ (30-3.447)e^.10x3 = 35.84

This is an Arithmetic series:

First term =2

Common difference = 2

Number of terms =37

Use Arithmetic Series formula to sum them up:

Sum =1,406