Ehh, now it's letting me answer.
So the associative property is basically as long as it's addition or multiplication, you can move around the groupings and it will still be the same.
For example, you can say:(21 * 7) (3x) instead of (21)(7)(3x)
And it would be equal to the old version.
(To answer your other question, the commutative property is you can move around all the parts of the equation as long as it's addition or multiplication.)
So basically, it would be the first one you sent me.
If a2 -2ab + b2 = 9 and a![\begin{gathered} a^2-2ab+b^2=9 \\ aStep 1 factorize[tex]\begin{gathered} a^2-2ab+b^2=(a-b)^2 \\ \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20a%5E2-2ab%2Bb%5E2%3D9%20%5C%5C%20a%3C%2Fp%3E%3Cp%3E%EF%BB%BFStep%201%20%3C%2Fp%3E%3Cp%3Efactorize%3C%2Fp%3E%5Btex%5D%5Cbegin%7Bgathered%7D%20a%5E2-2ab%2Bb%5E2%3D%28a-b%29%5E2%20%5C%5C%20%20%5Cend%7Bgathered%7D)
then
[tex]\begin{gathered} (a-b)^2=9 \\ \sqrt{(a-b)^2}=\sqrt{9} \\ a-b=\pm3 \\ \\ aa-b=-3
Answer:
(A)increasing
Step-by-step explanation:
Given the time and total distance ran by Myra in the table below.
We observe that as time increases, Myra's distance also increases. It is the speed(0.2 miles per minute) which is constant.
Therefore Option A is the correct option.
Answer:
Cost of each daylilies = $3
Cost of each ivy pot = $3
Step-by-step explanation:
Given:
Cost of 4 daylilies and 4 ivy = $24
Cost of 4 daylilies and 8 ivy = $36
Find:
Cost of each daylilies
Cost of each ivy pot
Computation:
Assume;
Cost of each daylilies = a
Cost of each ivy pot = b
So,
4a + 4b = 24....... eq 1
4a + 8b = 36 ....... eq 2
Eq2 - Eq1
8b - 4b = 36 - 24
4b = 12
b = 3
So,
Cost of each ivy pot = $3
4a + 4b = 24
4a + 4(3) = 24
4a + 12
a = 3
Cost of each daylilies = $3
