The student made the mistake in Step 2.
It should be m∠o + m∠p = 180 (Supplementary angles)
If this is corrected, then, since m∠m + m∠n + m∠o = 180 from Step 1,
we have,
m∠m + m∠n + m∠o = m∠o + m∠p which is Step 3.
Cancelling m∠o both sides, we get,
m∠m + m∠n = m∠p which is Step 4.
Answer:
2 11/12
Step-by-step explanation:
(1/3)×(35/4=35/12=2 11/12 yards
Answer:
<h2>n = 8</h2>
Step-by-step explanation:
Given the nth term of an arithmetic sequence to be Tn = a+(n-1)d
a = first term of the sequence
n = number of terms
d = common difference.
Given the first element a = 2 and 22nd to be 14
T22 = a+(22-1)d = 14
a+21d = 14
Substtuting a = 2 into the equation to get d
2+21d = 14
21d = 12
d = 12/21
d = 4/7
The nth term of the sequence given a = 2 and d = 4/7 will be expressed as;
Tn = 2+(n-1)4/7
Given Tn = 6
6 = 2+(n-1)4/7
6 = 2+4/7 n - 4/7
6-2+4/7 = 4/7 n
32/7=4/7 n
32 = 4n
n = 32/4
n = 8
Answer:
72 grams
Step-by-step explanation:
7.2 x 10 = 72
Let the lengths of the sides of the rectangle be x and y. Then A(Area) = xy and 2(x+y)=300. You can use substitution to make one equation that gives A in terms of either x or y instead of both.
2(x+y) = 300
x+y = 150
y = 150-x
A=x(150-x) <--(substitution)
The resulting equation is a quadratic equation that is concave down, so it has an absolute maximum. The x value of this maximum is going to be halfway between the zeroes of the function. The zeroes of the function can be found by setting A equal to 0:
0=x(150-x)
x=0, 150
So halfway between the zeroes is 75. Plug this into the quadratic equation to find the maximum area.
A=75(150-75)
A=75*75
A=5625
So the maximum area that can be enclosed is 5625 square feet.
