I need a test checker

[ARHQA$Y$YW4AYG4y]

So, I did a maths test today. The professor allowed us to take the tasks home so I decided to upload them here since we're not getting the test untill Tuesday.If you could calulate these and give me the results so I can know what I'm on...

Task #1: Calculate:

a) (-2^mbⁿ)⁴

b) a^-2+a^-3 ; if a=2

Task #2: Divide on factors:

a) a⁴b²-c²

Task #3: Shorten the fraction:

a³ - b³___
a²+ab+b²

Task #4: Calculate:
(the underline is the main fraction since it has double fractions)

x/a - x/2a____
x²/a² - x²/2a²

@Alen
@iverson954360

[Chester Bennington]

Work it out for yourself and you'll feel much better for it.

[Alen]

(-2^m b^n )^4=((-1)⋅2^m⋅b^n )^4=(-1)^4 (2^m )^4 (b^n )^4=2^4m b^4n=16^m b^4n
a^(-2)+a^(-3)=a⋅a^(-3)+1⋅a^(-3)=(a+1) a^(-3)=(2+1)/2^3 =3/8
a^4 b^2-c^2=(a^2 b)^2-c^2=(a^2 b-c)(a^2 b+c)
(a^3-b^3)/(a^2+ab+b^2 )=((a-b)(a^2+ab+b^2))/(a^2+ab+b^2 )=a-b
(x/a-x/2a)/(x^2/a^2 -x^2/(2a^2 ))=(x/2a)/(x^2/(2a^2 ))=(2a^2 x)/(2ax^2 )=a/x
Or with le substitution (most likely what the prof wanted):
(x/a-x/2a)/(x^2/a^2 -x^2/(2a^2 ))=(l-l/2)/(l^2-l^2/2),l=x/a
((l-l/2)⋅2)/((l^2-l^2/2)⋅2)=(2l-l)/(2l^2-l^2 )=l/l^2 =1/l=a/x

[ARHQA$Y$YW4AYG4y]

(-2^m b^n )^4=((-1)⋅2^m⋅b^n )^4=(-1)^4 (2^m )^4 (b^n )^4=2^4m b^4n=16^m b^4n
a^(-2)+a^(-3)=a⋅a^(-3)+1⋅a^(-3)=(a+1) a^(-3)=(2+1)/2^3 =3/8
a^4 b^2-c^2=(a^2 b)^2-c^2=(a^2 b-c)(a^2 b+c)
(a^3-b^3)/(a^2+ab+b^2 )=((a-b)(a^2+ab+b^2))/(a^2+ab+b^2 )=a-b
(x/a-x/2a)/(x^2/a^2 -x^2/(2a^2 ))=(x/2a)/(x^2/(2a^2 ))=(2a^2 x)/(2ax^2 )=a/x
Or with le substitution (most likely what the prof wanted):
(x/a-x/2a)/(x^2/a^2 -x^2/(2a^2 ))=(l-l/2)/(l^2-l^2/2),l=x/a
((l-l/2)⋅2)/((l^2-l^2/2)⋅2)=(2l-l)/(2l^2-l^2 )=l/l^2 =1/l=a/x

OMFG I'm getting an F...
I only solved 2/5

Now my GPA is 4.2 or something like that

[[MPGH]iverson954360]

This exam is not hard at all, you just need to practice your laws of exponents as well as a few rules pertaining to polynomial functions. Alen made solving the second one look harder than it needed to be. When you see something raised to a negative power you must put it on the opposite side of the fraction so in this case if you have all whole numbers over an a 1 ie. 2/1, 3/1, etc. then when a negative exponent comes into play you just flip the fraction ie (2^-2)/1 goes to 1/(2^2) which you then just solve simply as 1/4.

[Alen]

This exam is not hard at all, you just need to practice your laws of exponents as well as a few rules pertaining to polynomial functions. Alen made solving the second one look harder than it needed to be. When you see something raised to a negative power you must put it on the opposite side of the fraction so in this case if you have all whole numbers over an a 1 ie. 2/1, 3/1, etc. then when a negative exponent comes into play you just flip the fraction ie (2^-2)/1 goes to 1/(2^2) which you then just solve simply as 1/4.

I broke it up in a manner that I though would make it easiest to understand what I did Easiest way to "solve" it would be to simply solve

[ARHQA$Y$YW4AYG4y]

@Alen the second one where your result was 3/8, couldn't I just went with:

a^-2+a^-3 ; if a=2

2^-2 + 2^-3=
4 + (-8)= -4?

[Alen]

@Alen the second one where your result was 3/8, couldn't I just went with:

a^-2+a^-3 ; if a=2

2^-2 + 2^-3=
-4 + 8 = 4?

Nope.

[ARHQA$Y$YW4AYG4y]

@Alen
So dead...

Another question...

Shouldn't the tast go like this?

x/a - x/2a
x²/a² - x²/2a²=

2x-x/2a
2x²-x²/2a²=

x/2a
x²/2a²=

2a²x/2ax²=

2a/x

[Alen]

@Alen2a²x/2ax²=

2a/x

Nope.

[ARHQA$Y$YW4AYG4y]

OMFG!!!!!!