All Answers 358087 Answers

it is $4

I must concur with rachie, is there an interest rate? Or perhaps a fee Bob must pay in an alloted time? If none of these factors apply then the amount of currency Bob is in possesion of is 4 U.S.D.

Assumng that we are talking about rational functions, it is not possible for a function to have a horizontal asymptote and an oblique asymptote at the same time. The reason for this is that a horizontal asymptote will occur when we have the same degree polynomial in the numerator as in the denominator. A slant asymptote will occur when the degree of the polynomial in the numerator is greater than the degree of the polynomial in the denominator.......thus we can have the following combo possibilities :

vertical asymptote / horizontal asyptote  ....example      [ x+ 3] / [x - 1]

vertical asymptote / slant asymptote  ...... example    [ x^2] / [ x - 1]

.0000000127251967%

Whats the interest rate again?

If non $4. Trick Question?

I second 123,456,787,654,321 

n/6+3=12

5. (x^2y^4)^4/(y^-2)^4 =

x^8 * y^16 / y^-8

But remember....a negative exponent "throws" the result back  across the fraction bar  and makes the exponent positive........so we have

x^8 * y^16 * y^8   =

x^8 * y^24

6. (16x^2y^-4z^-1)^0/y^2 =   1 / y^2      [assuming x, y and z are not = 0 ]

7. (3ab^-1)^2  =

3^2 * a^2 * b^-2  =

9a^2 / b^2

Thee art most welcome.  :))

thank you ma-am 

Answered :))

Good day to you Mistress Katie-Leigh.  I heard the keys on your tunic jangling and I came a running.   :)

When the battle began, there were 30 percent more brigantines than men-of-war. If there were 260 brigantines, how many total soldiers took part in the battle?

Let the number of Brigantines be B   and the number of men-at-war be  M

and

B = 260

100% of M + 30% of M = B

1M              + 0.3M        =260

              1.3M                =260

I could have got to this point easier just by saying that 130% of M =260

Divide both sides by 1.3 and you get    M = 260/1.3

260/1.3 = 200

there were 200 men-at-war.    :)

Check

200+30% of 200 = 260  correct   

11111111^2 = 11111111 * 11111111 = 123456787654321

You think, but I think that is beyond your mental capacities.

28 inches / 100 = 7 / 25 inches or .28 inches

Good day to you High School Calculus,

I thought that your aswer was quite good.  You gave some indication of what the problem was without giving full answers.

I encourage responses like this.  We give full answers far too often.

In this case I could really see from previous responses that Copead was really trying hard to learn from us so I decided to elaborate.

BUT

Like I said I thought your answer was good which why I gave you 5 stars.  :)

Think about it this way:

Let x=0.333333 etc. Then, 10x=3.33.... Subtract x from both sides of the equation. The result is that 9x=3. Divide by 9 on both sides of the eqation, and you get x=3/9, or x=1/3. So, yes, technically, .333... multiplied three times will not equal exactly 1, but algebraicaly, it can be proven that this is the decimal form of 1/3, and that 3/3 thus equals 1.

Thanks for explaining everything Melody, I would have but my lazy bone acted up. Also Chemistry was calling my name.

1. (3x^4)^2 = 6x^8      wrong as explained by     HighSchoolCalculus.

2. (3x^2y^4)(-2y^-8) = 9xy^4256y    

I will work this one through slowly for you then maybew you can present the answers again and someone will check them for you :)

\((3x^2y^4)(-2y^{-8})\\ =3x^2y^4\times-2y^{-8}\\ =3\times x^2\times y^4\times-2\times y^{-8}\\ =3\times-2\times x^2\times y^4\times y^{-8}\\ =-6\times x^2\times y^{4+-8}\\ =-6\times x^2\times y^{-4}\\ =\frac{-6\times x^2\times y^{-4}}{1}\\ =\frac{-6\times x^2}{y^{+4}}\\ =\frac{-6x^2}{y^{4}}\\ \)

3. -4/x^-6 = -4x^6    correct

4. (x^-2y^-4)^-3(y^4y^3) = x^6y^19    correct

\( (x^{-2}y^{-4})^{-3}(y^4y^3) = x^6y^{19}     \)

5. (x^2y^4)^4/(y^-2)^4 = x^6y^2

\((x^2y^4)^4/(y^{-2})^4 = x^6y^2\qquad wrong \)

? 6. (16x^2y^-4z^-1)^0/y^2 = 3/y^2 ?    wrong    -     Anything (except 0) raised to the power of 0 is 1

a = -2 b=6       huh?

7. (3ab^-1)^2 = 9a^2b       wrong

? 8.(4a^2b^-4)^0 ?

9.(x^5)^0=1      correct

?10.(a^2b^3)(missing exponent) = 1/a^2b^3 ?

11. x^-5*x^11=x^6       correct

12.w^8/w^5 = w^3       correct

Actually the answer is 42. Don't question it.