625^-3/4 i know this equals 1/125

for (-32)^3/5 this equals -8 right? I was wondering if someone could explain to me what happens to the 1 and why it is not needed in the answer, but compared to the top one it is needed.

first, yes (-32)^(3/5) = -8

the difference between the two is the negative sign in the exponent.

if you ignore the negative sign at first, you get 125^(-1)...

In other words 625^( 3/4) is 125.

Any number to a -1 power will be equal to one divided by the number. Its a standard exponent property.

The first question, the exponent is negative, according to the exponent rule you have to put it under 1 as a fraction.

Second question, the exponent is positive therefore no fraction is needed.

we know that 10^-2= 10^2 so similarly:

625^-3/4 = 1/ (625^3/4)

obviously 5^4=625 thus

1/(625^3/4) = 1/(5^4^(3/4))

we know 2^5^3= 2^(5*3) = 2^15, similarly

1/(5^4^(3/4) = 1/5^(4*3/4) = 1/5^3

5^3 = 125 therefore

1/5^3 = 1/125

proved

cheers

## Thursday, July 22, 2010

### CALCULUS- sketch a graph of ONE function that satisfies all of the given conditions?

so my teacher gave us the following problem.. and I don

### What is the remainder when 9^6 - 11 divided by 8?

plz b fast.......i stuck up at this question....

9 = 1 (mod 8)

right? Well, then:

9^6 = 9 * 9 * 9 * 9 * 9 * 9 = 1 * 1 * 1 * 1 * 1 * 1 = 1 (mod 8)

So:

9^6 - 11 = 1 - 3 = 6 (mod 8)

(9^6-11) / 8

=(531,441-11) / 8

=531,430 / 8

=66428.75

=

9 = 1 (mod 8)

right? Well, then:

9^6 = 9 * 9 * 9 * 9 * 9 * 9 = 1 * 1 * 1 * 1 * 1 * 1 = 1 (mod 8)

So:

9^6 - 11 = 1 - 3 = 6 (mod 8)

(9^6-11) / 8

=(531,441-11) / 8

=531,430 / 8

=66428.75

=

### I just took 250 mg of Tylonol Simply Sleep, am I okay?

How much is deadly?

Well, lets find out shall we?

Given that they are the 25 mg tablets, that means you took 8 more pills than you should have. That

Well, lets find out shall we?

Given that they are the 25 mg tablets, that means you took 8 more pills than you should have. That

### Statistics help whats the difference between a t distribution and a z score?

Is it true you only use a t distribution when n is less than 30

Yes, but under some certain circumstances. We can use a t distribution when we know the distribution of the variable is normal, so we use t distribution (this is a condition isn

Yes, but under some certain circumstances. We can use a t distribution when we know the distribution of the variable is normal, so we use t distribution (this is a condition isn

### Why doesn

So according to vertical motion, you add the accelaration to the initial velocity forever. Therefore, if the initial velocity is 6 and accelaration is -1, it should be:

6, 5, 4, 3, 2, 1, 0 (and that would make a total of 21) every second

but when I use the formula, d = -u^2/2a I get 18 instead? Why is this happening?

From 0 to 1 sec, the average velocity is (6 5)/2 = 5.5

The second second, the average is 4.5.

Then 3.5, 2.5, 1.5, .5

Add them up and you get 18.

Because acceleration doesn

6, 5, 4, 3, 2, 1, 0 (and that would make a total of 21) every second

but when I use the formula, d = -u^2/2a I get 18 instead? Why is this happening?

From 0 to 1 sec, the average velocity is (6 5)/2 = 5.5

The second second, the average is 4.5.

Then 3.5, 2.5, 1.5, .5

Add them up and you get 18.

Because acceleration doesn

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