My answer -
<span>1. Use symbols (not words) to express quotient
2. Use exponent symbol (^) to denote exponents
3. Just write out question number, question, and choices. No need for
extra information (such as points). Also, don't leave blank lines
between choices. This extraneous that we don't need just makes your
whole question very very long, and means a lot of scrolling on our part.
4. You should only post 2 or 3 questions at a time.
1) (6x^3 − 18x^2 − 12x) / (−6x) = −x^2 + 3x + 2 ----> so much simpler to read !
2) (d^7 g^13) / (d^2 g^7) = d^(7−2) g^(13−7) = d^5 g^6 ----> much easier to read !
3) (4x − 6)^2 = 16x^2 − 24x − 24x + 36 = 16x^2 − 48x + 36
4) (x^2 / y^5)^4 = (x^2)^4 / (y^5)^4 = x^8 / y^20
5) (3x + 5y)(4x − 3y) = 12x^2 − 9xy + 20xy − 15y^2 = 12x^2 + 11xy − 15y^2
6) (3x^3y^4z^4)(2x^3y^4z^2) = (3*2) x^(3+3) y^(4+4) z^(4+2) = 6 x^6 y^8 z^6
7) 5x + 3x^4 − 7x^3 ----> Fourth degree trinomial
8) (5x^3 − 5x − 8) + (2x^3 + 4x + 2) = 7x^3 − x − 6
9) (x − 1) + (2x + 5) − (x + 3) = x + 1
10) (−4g^8h^5k^2)0(hk^2)^2 = 0 (anything multiplied by 0 = 0)
or.. (−4g^8h^5k^2)^0(hk^2)^2 = 1 (h^2 (k^2)^2) = h^2 k^4
Last question shows why it is so important to use proper symbols (such
as ^ to indicate exponents). Without such symbols, I could not tell if
the 0 was an actual number and part of multiplication, of if 0 was an
exponent of the expression preceding it.
P.S
Glad to help you have an AWESOME!!! day :)
</span>
Answer:
29.64
Step-by-step explanation:
First we multiply 0.3 with 22.8.This will give us 6.84 which is the 30% off.Now we add 22.80 and 6.84 and we will get our answer 29.64.
Answer:
Lin needs 3 cups of flour and 6 cups of sugar. Noah needs 6 cups of sugar and 4 cups of flour.
Step-by-step explanation:
I want you to figure out how to do.
Answer:
Second choice:
Fifth choice:
Step-by-step explanation:
Let's look at choice 1.
I'm going to subtract 1 on both sides for the first equation giving me 
. I will replace the 
in the second equation with this substitution from equation 1.
Expand using the distributive property and the identity 
:
So this not the desired result.
Let's look at choice 2.
Solve the first equation for by dividing both sides by 2:
.
Let's plug this into equation 2:
This is the desired result.
Choice 3:
Solve the first equation for by adding 3 on both sides:
.
Plug into second equation:
Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:
Not the desired result.
Choice 4:
I'm going to solve the bottom equation for since I don't want to deal with square roots.
Add 3 on both sides:
Divide both sides by 2:
Plug into equation 1:
This is not the desired result because the 
variable will be squared now instead of the 
variable.
Choice 5:
Solve the first equation for by subtracting 1 on both sides:
.
Plug into equation 2:
Distribute and use the binomial square identity used earlier:
.
This is the desired result.
