Answer:
student ticket = $11
(adult ticket = $15)
Step-by-step explanation:
Let a = price of adult ticket
Let s = price of student ticket
Given:
- On the first night she sold 12 adult tickets and 11 student tickets for $301 dollars
⇒ 12a + 11s = 301
Given:
- On the second night she made $134 selling 6 adult tickets and 4 student tickets
⇒ 6a + 4s = 134
Multiply 6a + 4s + 134 by 2 then subtract from 12a + 11s = 301 to eliminate a:
⇒ (6a + 4s = 134) × 2: 12a + 8s = 268
12a + 11s = 301
- (12a + 8s = 268)
--------------------------
3s = 33
⇒ s = 33 ÷ 3 = 11
Substitute found value of s into one of the equations and solve for a:
⇒ 12a + 11(11) = 301
⇒ 12a + 121 = 301
⇒ 12a = 180
⇒ a = 15
Therefore, the price of an adult ticket is $15 and the price of a student ticket is $11
Answer:
Option D
Step-by-step explanation:
Option A
x² - 3x² + 2x² = 0
Therefore, the given polynomial has no roots.
Option B
x² + 2x + 8
By quadratic formula,
x = 
= 
Therefore, roots are imaginary and for a polynomial with degree 1 or more than 1, both the roots can't be imaginary.
So the polynomial can't have exactly two roots.
Option C
99x³ - 33x + 1
Since the polynomial is of degree 3, so it will have three roots.
Therefore, the polynomial will not have exactly 2 roots.
Option D
√2x - 3x² + 7√2
By quadratic formula,
x = 
x = 
There are exactly two real roots.
Therefore Option D is the answer.
Option E
4x + 11x - 111
Since this polynomial is of a degree 1.
There will be only one root.
Cos(∅) = adjacent ÷ hypotenuse
cos(39) = x ÷ 17m
cos(39) ≈ 0.77715 = x ÷ 17m
x = (0.77715)(17m)
x ≈ 13.2114
Answer:
2+1p^2/2
Step-by-step explanation:
Good morning,
Answer:
the lower class limit of the class [13,17[ is 13.
:)
