MATH TIME!

Just re-starting math after 3 years so getting the basics back! @MarkFL


Evaluate the variable expression for x = 1, y = −2, and z = −3.
xy − z(x − y)2

Would it be (1 x -2) - (-3)(1-(-2))^2? Then solve? the fun stuff starts later T.T

 

Yes, the first step is to plug in the given numbers for the variables, as you've done. Then recall the order of operations. I would write this as:

(1)(-2) - (-3)((1) - (-2))^2

I would work within the parens first:

(1)(-2) - (-3)(1 + 2)^2

(1)(-2) - (-3)(3)^2

Now the exponent:

(1)(-2) - (-3)(9)

Now the multiplications:

-2 - (-27)

-2 + 27

25

Your calculator may be able to evaluate that directly (to check your work), or you can use W|A:

(1)(-2) - (-3)((1) - (-2))^2

 

@MarkFL Thank you!!!!! I'll be getting some more for you soon! haha

 

I wish there was a law equivalent of you

 

@MarkFL is a true hero
I start back to school on the 15th and will be taking math can I ask for your help in advance?

 

Certainly! I will be glad to help.

 

Why don't you be that guy, lol?

 

Thank you uncle mark

 

I'll just change my course to maths

 

Keep at it bro we got this

 

To announce, this was the correct answer !

 

Oh yeah!

 

What's:

(7x6) SU (45)= CK + 12 / MY (67)+ DI (65) - CK?

 

That equation has no solution over the reals.

 

Yes, it does

 

(a^5b/a^3b^-19)^2

@MarkFL I keep getting b^16/a^4

 

sorry b^-9***

 

Okay, I would write it as:

[(a^5b)/(a^3b^(-9))]^2

I would first work within the outer brackets, using the rule for exponents:

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

to get:

[a^(5 - 3)b^(1-(-9))]^2

[a^2b^10]^2

Now square, using the rule:

(x^a)^b = x^(ab)

to get:

a^(2*2)b^(10*2)

a^4b^20

Does that make sense?

 

@MarkFL

I see I skipped to far ahead damn

 

It's gotten to be bedtime for Bonzo here, but I will be online in the morning.