CPhill

Here's a way to do it, but it's a "two-step" process

Say we have   7/4

1. Hit the enter key and you get  1.75   (remember the "whole part")

2. Subtract  the  "whole part" of the answer from 1.75 and the calculator will return the "mixed part" (fractional part)

how to solve for questions like 5-e^2x=-15

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I'm going to assume that you're asking this:

5 - e^(2x) = -15         .......subtract 5 from both sides and then multiply both sides by (-1)

-e^2x = -20

 e^(2x) = 20                 Now, take the "natural log," (ln), of both sides

ln e^(2x) =  ln 20          By a property of logs, the "2x" can be "brought out front" on the left side

(2x) lne = ln 20          and "lne" just equals 1     so we have

2x = ln 20                  so.......

x = ln 20 / 2  ≈  1.5

what is log base 3 to the 9th

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You are asking this question...... ( log₃ )⁹  ........but that has no meaning

Do you mean this ???

log₃ 9

If so, then we are asking, "What power (exponent) do we need to put on the base (3) to raise it to 9?"

So we have, in exponential form,

3^x = 9

And x = 2

I believe this is what you are asking........

An artist creates models of square pyramids. For each pyramid created, the height is 2 times the side length of the base, s. The cost of each pyramid is based on volume and is $0.30 per cubic centimeter.

The cost, C(s), in dollars of a pyramid with base side length, s, is given with the following function. 

Which statement best describes the coefficient of s³?

• the product of the ratio of the height to the side length and the price per cubic centimeter
  
•  

  the product of the ratio of the height to the base area, one-third, and the price per cubic centimeter
  
•  

  the product of the ratio of the height to the side length, one-third, and the price per cubic centimeter
  
•  

  the product of the ratio of the height to the base area and the price per cubic centimeter

   

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  The volume of a square pyramid in our example is  given by  V = (1/3)(s ²)(h)   Where s is the side length of the base (thus s ² is the area of the base) and h is the height.  In our example the height is twice the side length so we would have  V = (1/3)(s ²)(2s). And to get the total cost, we would multiply this volume by (.30). So we would have (1/3)(2)(.30)s ³ = (1/3)(2)(30/100)s ³ = (2/3)(30/100)s ³ = (20/100)s ³ = 0.2s ³ and the coefficient would be equal to the product of the ratio of the height to the side length (2), times one-third, times the cost per cubic centimeter (.30 or 30/100).

   

  Thus, the third choice is the correct one.

   

        

What does y=2 means?

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You might be refering to the graph of a line whose equation is y = 2. It's just a line parallel to the x axis that intercepts the y axis at the point (0, 2). Every point on the line has a y coordinate value of "2," no matter what its x coordinate value is.

Could you give us a specific example of what you're having trouble with?? In general, we can only reduce them when the odd number divides the even. An even can never divide an odd.

If you go to the Formulary Tab at the top of the page, you'll find a "cylinder" app that will calculate this for you  

Does 1/2x + 4y = 0 represent a direct variation ?

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Manipulating this a bit, we have

y = -(1/8)x

Notice that this is just a line with a negative slope that passes through the origin.....as x increases, y decreases

Thus we have an inverse variation.

2x+y=6 3x+y=2

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I assume you want to solve this system

2x+y=6

3x+y=2    

Let's just subtract the second equation from the first   We have

-x = 4     So

x = -4  

And putting this value back into the second equation, we find that

3(-4) + y = 2

-12 + y = 2     Add 12 to both sides

y = 14

So x = -4 and y = 14

You should check that these are correct by substituting them back into the first equation.

cosx =1/9 what is the value of cos 1/2x

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Using the identity cos [x/2] = ±SQRT [(1 +cos(x) / 2] = ±SQRT [(1 + (1/9) / 2] = ±SQRT [5/9] =

±SQRT (5) / (3)

If x lies in the first quadrant, take the positive root; if x  lies in the fourth quadrant, take the negative root.