Pre Algebra

[cbrenda]

please solve these two problem

 

[tkhunny]

Okay, I got them without much problem. How did you do?

 

[mmm4444bot]

Remember the meaning of exponents; they tell us how many factors there are.

EG:

A^5/A^7 = (A*A*A*A*A)/(A*A*A*A*A*A*A)

In other words, A^5/A^7 means five factors of A in the numerator and seven factors of A in the denominator.

Simplifying means canceling all five factors of A on top with five of the seven factors of A on bottom.

1/(AA)

We're left with 1/A^2.

By the way, we could let A = -2

Now, what do you think about your first exercise?

PS: Your work seems to show that you've mixed up two properties of exponents.

If you were to use the property for dealing with ratios of powers, then you should subtract the lower exponent from the top exponent. Here is that property:

n^a/n^b = n^(a - b)

We only add exponents when dealing with products:

(n^a) * (n^b) = n^(a + b)



Here's an example for your second exercise.

Percent of change = (new value - old value)/(old value)

In other words, we divide the difference between values by the old value.

EG:

A computer price increases from $1,200 to $1,500. What is the percent of change?

Substitute the known values into the formula above, and do the arithmetic.

The old value is 1200

The new value is 1500

Percent of change = (1500 - 1200)/1200 = 1/4 = 0.25

The percent of change is 25%.

Can you try something similar?

 

[mmm4444bot]

It's been awhile; I hope that you were able to finish these two exercises.

Here's my work, for future reference.

---------------------------------------------------------

(-2)^5/(-2)^7

(-2)^(5 - 7)

(-2)^(-2)

The first answer is (A): 1/(-2)^2

---------------------------------------------------------

Old value: 50

New value: 400

Percent of change = (400 - 50)/50

= 350/50

= 7

The second answer is (D): 700%

 

[lookagain]

please solve these two problems

cbrenda,

we are not here to "solve" your (math) problems.
Show your work/attempts first.

You can reasonably state "Please give me guidance on these problems
(as you can see my attempts here so far)." Or, it may be a variation
on that.

 

[polygonjordy]

please solve these two problems

cbrenda,

we are not here to "solve" your (math) problems.
Show your work/attempts first.

You can reasonably state "Please give me guidance on these problems
(as you can see my attempts here so far)." Or, it may be a variation
on that.

This is exactly right. Tell us a little background in to what exactly you are trying to understand.