S= one length of square 1
S+1 = one length square 2
S+2= one length square 3
Our equation would be
S^2+(S+1)^2+(S+2)^2=365
Idk if it would be called distribute but.. distribute
S^2+S^2+2S+1+S^2+4S+4=365
Add like terms
3S^2+6S-360=0
Divide everything. By 3
S^2+2S-120=0
Factor
(S-10)(S+12)=0
The two solutions are:
S-10=0
S=10
S+12=0
S=-12
Since a length can’t be a negative the only possible solution would be 10
Since a perimeter is all lengths added together we can multiply the length by four to get the perimeter
Square 1
10*4=40
Perimeter is 40cm
Square 2
S+1 =11
11*4=44
Perimeter is 44cm
Square 3
S+2=12
12*4=48
Perimeter is 48cm
Add all the perimeters together to get the total perimeter:
Total perimeter:
40+44+48=132
The total perimeter is 132cm
I hope this helps. Sorry if I messed up anything on here it was kinda hard to keep track of everything. Feel free to ask if you need anything cleared up :)
Answer:
- The mean for Group A is less than the mean for Group B.
- The median for Group A is less than the median for Group B.
- The mode for Group A is less than the mode for Group B.
Step-by-step explanation:
First, we can find the measures of center for each group.
<u>Group A</u>
Mode: 1
Median: (1 + 2) / 2 = 3 / 2 = 1.5
Mean: (1 * 5 + 2 * 4 + 3) / 10 = (5 + 8 + 3) / 10 = 16 / 10 = 1.6
<u>Group B</u>
Mode: 3
Median: 92 + 3) / 2 = 5 / 2 = 2.5
Mean: (1 * 3 + 2 * 2 + 3 * 4 + 5) / 10 = (3 + 4 + 12 + 5) / 10 = 24 / 10 = 2.4
From here, we can see that...
- The mean for Group A is less than the mean for Group B.
- The median for Group A is less than the median for Group B.
- The mode for Group A is less than the mode for Group B.
Hope this helps!
QUESTION 1
Given that:
,
,
and
Then;
Group similar terms;
Simplify;
QUESTION 2
Given that;
.
and
Substitute the functions;
Substitute x=3
QUESTION 3
Given:
This implies that;
Expand the parenthesis;
QUESTION 4
The given function is;
Let
The range is:
The interval notation is;
