Since you know the area and the width, you can divide the area by the width to find the length. This gives you a length of 110 yards. To find the perimeter of the soccer field, you would add all the side lengths together. Since the width is 70, you would add 70 + 70, along with the length, 110 + 110. This gives you a perimeter of 360.
Given :
Raffle tickets were sold for a school fundraiser to parents, teachers, and students. 563 tickets were sold to teachers. 888 more tickets were sold to students than to teachers. 904 tickets were sold to parents.
To Find :
How many tickets were sold to students.
Solution :
Ticket sold to teachers, T = 563 .
Ticket sold to parents, P = 904 .
Let, ticket sold to students are S.
Now, it is given that :
S = T + 904
S = 563 + 904
S = 1467 students
Therefore, tickets sold to students are 1467 .
Hence, this is the required solution.
Let p(x) be a polynomial, and suppose that a is any real
number. Prove that
lim x→a p(x) = p(a) .
Solution. Notice that
2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .
So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial
long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x –
2.
Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number
such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| <
1, so −2 < x < 0. In particular |x| < 2. So
|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|
= 2|x|^3 + 5|x|^2 + |x| + 2
< 2(2)^3 + 5(2)^2 + (2) + 2
= 40
Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2
+ x − 2| < ε/40 · 40 = ε.
Answer:
$240.5
Step-by-step explanation:
Multiply 0.85 and 130 to get 110.5. Add this to the original price, 130, to get 240.5.
if the distance from R to S is only half then the other half is S to T. The distance between R and S is (5,2). that should give you the answer
