Melody

If you multiply by a number bigger than 1 then then answer is bigger than the original.

eg

4*1.5 = 6 which is bigger than 4

also

An enlargement using a negative scale factor is similar to an enlargement using a positive scale factor, but this time the image is on the other side of the centre of enlargement, and it is upside down.

SO is this enough infromation for you to answer your own question?

Contractions are smaller by the way.

Thanks Alan,

Another for me to study  

All studied.

Equating coefficients - maybe I will think of it myself next time.   Thanks once again :)

Это лучший способ сделать это.

This is a better way to do it.

\(\displaystyle\lim_{x\rightarrow 2}\;\frac{x^2-x-2}{x^2-3x+2}\\ =\displaystyle\lim_{x\rightarrow 2}\;\frac{(x-2)(x+1)}{(x-2)(x-1)}\\ =\displaystyle\lim_{x\rightarrow 2}\;\frac{(x+1)}{(x-1)}\\ =\frac{3}{1}\\ =3\)

7.1

It is important to note that both the numerator and the denominator tend to 0 so I will use L'Hopital's rule.

\(\displaystyle\lim_{x\rightarrow 2}\;\frac{x^2-x-2}{x^2-3x+2}\\ =\displaystyle\lim_{x\rightarrow 2}\;\frac{\frac{d}{dx}(x^2-x-2)}{\frac{d}{dx}(x^2-3x+2)}\qquad \text{Using L'hopital's Rule }\\ =\displaystyle\lim_{x\rightarrow 2}\;\frac{2x-1}{2x-3}\\ =\frac{3}{1}\\ =3\)

The diameter of a circle is 16. By what number must the radius be decreased in order to decrease the area of the circle by 48 pi?

I really am not sure what you are asking.....

The original area is   pi*64    if you want to decrease this by 48pi  then the new area must be 16pi units^2

So the radius would have to be 4 units, the diameter would be 8 units.

This is half of the orginal diameter. 

Hello Rosala, it is good to see you :)

I have played with this question for quite a while and it is driving me nuts  LOL   

I'd be interested to see someone else answer this or if you get an answer at college I would like to see that too :)

Hi UndertaleButtercup32

It is good that you solved them yourself and also good that you told us that you no longer needed help but please leave the actual question alone next time. I mean don't strike it out because some people .. like ProMagma may still like to play with it :)

Thanks ProMagma :)

ProMagma,

Could you please tell me what the formula for the volume of a sphere is?

You can look it up, that is ok :)