A- price of adult's ticket
c- price of children's ticket
a+3c=20
a=c+2
c=a-2
a+3(a-2)=20
a+3a-6=20
4a=26
a=6.5
The cost of an adult's ticket is $6.50
2x+2y=16
<span>2x+6y=28
all you have to do is subtract to get rid of the 2x and solve for y
2x+2y = 16
-2x-6y = -28
---------------------------
0 -4y = -12
y = 3
now we got y we can plug in or do the systems of equation again and find x. lets do systems of equations again.
</span>2x+2y=16
<span>2x+6y=28
</span><span>
this time i am going to multiple the top by -3 and then add the equations hopefully it is clear why I multiple by -3
</span><span>
-3(2x+2y = 16)
2x +6y = 28
-6x - 6y = -48
</span>2x +6y = 28
------------------------
-4x + 0y = -20
x = 5
Answer:
B. 14x = 42
Step-by-step explanation:
The figure is composed of a rectangle and a trapezium
(L*W) + [½(a + b)h] = Area of the figure
Where,
Area = 42
L = 2x
W = 3
a = 3 + 4 = 7
b = 9
h = x
Plug in the values
(2x*3) + [½(7 + 9)x] = 42
6x + [½(16x)] = 42
6x + 8x = 42
14x = 42
Answer:
a)
in
b) 28 in
c) 784 in²
Step-by-step explanation:
Let the length be 'L'
and the radius be 'r'
Thus, according to the question
L + 2πr = 84 in
L = 84 - 2πr ............(1)
Volume of the cylinder, V = πr²L
substituting the value of L from 1, we get
V = πr²(84 - 2πr)
or
V = 84πr² - 2π²r³
for points of maxima, differentiating the above equation and equating it to zero

or
2(84)πr - 3(2)π²r² = 0
or
2πr(84 - 3πr) = 0
or
r = 0 and 84 - 3πr = 0
or
⇒ 3πr = 84
or
⇒ r =
in
since, the radius cannot be zero therefore, r = 0 is neglected
Therefore,
a) The radius of the largest cylindrical package =
in
b) from (2)
L = 84 - 2πr
or
⇒ L = 
or
⇒ L = 84 - 56 = 28 in
The length of the largest cylindrical package = 28 in
c ) The volume of the largest cylindrical package ,V = πr²L
= 
= 784 in²