Answer:
0 ≤ x ≤ 18
Step-by-step explanation:
Let 'x' be the number of youth tickets purchased at the zoo and 'y' be the number of adult tickets purchased at the zoo.
At a zoo, youth tickets cost $5 and adult tickets cost $9. A group spent a total of $90 on tickets. We can write as
5x + 9y = 90
to find x we divide both side by 5
x = (90-9y)/5 => 90/5 - 9y/5
x = 18 - 9y/5
The domain of the relationship is the possible set of values of x and y that satisfies the equation. The domain of this relationship is
0 ≤ x ≤ 18
At x = 0, it means only adult tickets were purchased.
At x = 18, it means only youth tickets were purchased.
The answer is C. 100 because the total measure of a quadrilateral is 360. When you add up the 3 interior angle measures, it adds up to 260. Then you subtract 360-260 to get 100 as the fourth angle measure. Hope that this helped!
Given that
(2/x)+(3/y) = 13--------(1)
(5/x)-(4/y) = -2 -------(2)
Put 1/x = a and 1/y = b then
2a + 3b = 13 ----------(3)
On multiplying with 5 then
10a +15 b = 65 -------(4)
and
5a -4b= -2 ----------(5)
On multiplying with 2 then
10 a - 8b = -4 -------(6)
On Subtracting (6) from (4) then
10a + 15b = 65
10a - 8b = -4
(-)
_____________
0 + 23 b = 69
______________
⇛ 23b = 69
⇛ b = 69/23
⇛ b =3
On Substituting the value of b in (5)
5a -4b= -2
⇛ 5a -4(3) = -2
⇛ 5a -12 = -2
⇛ 5a = -2+12
⇛ 5a = 10
⇛ a = 10/5
⇛ a = 2
Now we have
a = 2
⇛1/x = 2
⇛ x = 1/2
and
b = 3
⇛1/y = 3
⇛ y = 1/3
<u>Answer :-</u>The solution for the given problem is (1/2,1/3)
<u>Check</u>: If x = 1/2 and y = 1/3 then
LHS = (2/x)+(3/y)
= 2/(1/2)+3/(1/3)
= (2×2)+(3×3)
= 4+9
= 13
= RHS
LHS=RHS is true
and
LHS=(5/x)-(4/y)
⇛ 5/(1/2)- 4/(1/3)
⇛(5×2)-(4×3)
⇛ 10-12
⇛ -2
⇛RHS
LHS = RHS is true
C=2x Pi x R so 2x3.14x17.5 = 109.9
The two labeled angles are alternate interior angles, and as such, they are the same.
From this result you can build the equation
and solve it for x: subtract 13x from both sides to get
and add 2 to both sides to get
Check: if we plug the value we found we have
So the angles are actually the same, as requested.
