LCD = 12
Thanks (:
!!!!!!!!
I'm really sorry, I'm not the best at this, but I did all the equations the best to my knowledge and came up with these 3 answers
(2,-1) which equaled -7
(5,4) which equaled -11
and (0,-9) which equaled - 9
Again I'm really sorry if this didn't work. :( God bless and good luck with future lessons.
Answer:
Let l be the length and w the width of the rectangle.
P as the perimeter and A as the area.
length of a rectangle is four times the width:
w = w
l = 4w
System of equation : 2l + 2w = P
Substitute:
2(4w) + 2w = 45
8w + 2w
10w = 45
w = 45 / 10
w = 4.5
Solve for l:
l = 4w
l = 4(4.5)
l = 18
Area:
A = lw
A = 18(4.5)
A = 81 in²
Answer:
6
Step-by-step explanation:
The trick here is to know PEMDAS.
PEMDAS stands for: Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
This is the order in which you should simplify a problem.
Parentheses:
(5 + 1)^2 - (11 + 32) + 4
(6)^2 - (43) + 4
Exponents:
(6)^2 - (43) + 4
36 - 34 + 4
We have no multiplication or division, so we can skip those steps.
Addition:
36 - 34 + 4
40 - 34
Subtraction:
40 - 34
6
So 6 is our answer.
<h3>
Answer: Largest value is a = 9</h3>
===================================================
Work Shown:
b = 5
(2b)^2 = (2*5)^2 = 100
So we want the expression a^2+3b to be less than (2b)^2 = 100
We need to solve a^2 + 3b < 100 which turns into
a^2 + 3b < 100
a^2 + 3(5) < 100
a^2 + 15 < 100
after substituting in b = 5.
------------------
Let's isolate 'a'
a^2 + 15 < 100
a^2 < 100-15
a^2 < 85
a < sqrt(85)
a < 9.2195
'a' is an integer, so we round down to the nearest whole number to get 
So the greatest integer possible for 'a' is a = 9.
------------------
Check:
plug in a = 9 and b = 5
a^2 + 3b < 100
9^2 + 3(5) < 100
81 + 15 < 100
96 < 100 .... true statement
now try a = 10 and b = 5
a^2 + 3b < 100
10^2 + 3(5) < 100
100 + 15 < 100 ... you can probably already see the issue
115 < 100 ... this is false, so a = 10 doesn't work
