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It is used in trigonometry.

If you are interestested here is a you tube clip

2.5p + 1 = 1.4p

2.5p - 1.4p =-1

1.1p           = -1

p               = -1/1.1

                 = -10/11 =-0.90909090909090909090909090909091

so that p =- 0.9090//

Differentiate means 

Find the equation that will give the gradient of the tangent to the curve at any given point.  

Interesting question and great pic Chris  

2^x = 225    take the log of both sides

log 2^x = log 225   and we can write

xlog 2 = log 225       divide both sides by log 2

x = log225/log2 = 7.8137811912170377

0.15*35 pounds

And yet, here again is another possibility....if you have had calculus, we know that the derivative = 0 where the turning points occur,

Therefore, let the derivative of the function be (x -3) (x +2) = x^2 - x - 6.

Then, again.... this may only make sense if you have had calculus.......our basic function would be:

F(x) = (1/3)x^3 - (1/2)x^2 - 6x

This function will have a y intercept of 0, so we  just need to append a "10" at the end to give us a y intercept of 10.

So....our "translated" function would be...

F(x) = (1/3)x^3 - (1/2)x^2 - 6x + 10

multiply by 100 and put a percent sign

3=300%

5x+10y=220

post again with brackets please - your question is not clear.

110/3000*100 percent

that depends on the question :)

2.2/100*1925

The gradient is 4 and the y intercept of -7.  

Any line with a gradient of 4 (but with a different y intercept) wil be parallel.

You need to show how you get your answer Bio

Ohterwise how can any learning take place?

2.25=1.5%of x

2.25=0.015X

X=2.25/0.015

Yes, that works too.  I always find it easier to remember just the sine and cosine relationships though, and work out the other trig identities from those if I need them.

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Got it, thanks. I worked out i could have also used sec²x=1+tan²x and cosec²x=1+cot²x so that the 1s cancelled out. 

Ahh I see! Thankyou 

Again, use sines and cosines

1/cos²x - 1/sin²x = sin²x/cos²x - cos²x/sin²x

right-hand side can be written as: (sinx/cosx + cosx/sinx)(sinx/cosx - cosx/sinx) 

                                              = (sin²x + cos²x)/(sinxcosx)*(sinx/cosx - cosx/sinx) 

                                              = 1/(sinxcosx)*(sinx/cosx - cosx/sinx) 

                                              = 1/cos²x - 1/sin²x = right-hand side

NB:  You need to remember that sin²x + cos²x =  1.  This is an important identity.

Try changing to sines and cosines.

cosA/sinA - sinA/cosA = 2cos2A/sin2A

cos2A = cos²A - sin²A

sin2A = 2sinAcosA

so, right-hand side is 2cos2A/sin2A = 2(cos²A - sin²A)/(2sinAcosA) = cos²A/(sinAcosA) - sin²A/(sinAcosA) = cosA/sinA - sinA/cosA = left-hand side.

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$$60\times \log{(1+r/12)}=\log{1.8}$$

Divide both sides by 60

$$\log{(1+r/12)}=\frac{\log{1.8}}{60}$$

Assuming the log is to the base 10 then:

$$1+\frac{r}{12}=10^{\frac{\log{1.8}}{60}}$$

so:

$$r=12(10^{\frac{\log{1.8}}{60}}-1)$$

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Here's another possible approach:

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Hi Chris

I already gave that address.

I used a hyperlink.

Tenacious showed me how to do it 😀

2^6 is 64, so there would be 64 times as many at the end of the hour as at the start.

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