Answer:
a) The cost of ribeye steak dinners is $ 16.96.
b) The cost of grilled salmon dinners is $ 23.37.
Explanation:
Let the cost for ribeye steak dinners be m and cost for grilled salmon dinners be n.
A waitress sold 10 ribeye steak dinners and 18 grilled salmon dinners, totaling $590.39 on a particular day.
So, we have
10 m + 18 n = 590.39 -------equation 1
Another day she sold 22 ribeye steak dinners and 9 grilled salmon dinners, totaling $583.49
So, we have
22 m + 9 n = 583.49 -------equation 2
equation 2 x 2
44 m + 18 n = 1166.98 -------equation 3
equation 3 - equation 1
44 m + 18 n - ( 10 m + 18 n ) = 1166.98 - 590.39
m = 16.96 $
Substituting in equation 1
22 x 16.96 + 9 n = 583.49
n = 23.37 $
a) The cost of ribeye steak dinners is $ 16.96.
b) The cost of grilled salmon dinners is $ 23.37.
The remaining balance would be $3,600.
$6,300/7 = 900 x 3 = $2,700
$6,300 - $2,700 = $3,600
Answer:
Yes
Step-by-step explanation:
Answer:
3/2(x+y)
Step-by-step explanation:
The cost of 2 DVD is x+y
Divide by 2 to get the cost of one DVD
(x+y) /2
Multiply by 3 to get the cost of 3
3/2(x+y)
Refer to the diagram shown below.
When x = 30 ft, the cable is at 15 ft, therefore y(30) = 15.
That is,
a(30 - h)² + k = 15 (1)
Also, because the distance between the supports is 90 ft, therefore
y(0) = 6 ft, and y(90) = 6 ft
That is,
a(-h)² + k = 6 (2)
a(90 - h)² + k = 6 (3)
From (2) and (3), obtain
a(90 - h)² = ah²
90² - 180h + h² = h²
180h = 90²
h = 45 ft.
From (1) and (2), obtain
225a + k = 15
2025a + k = 6
Therefore
1800a = -9
a = - 0.005
k = 15 - 225(-0.005) = 16.125 ft
Answer:
The equation for the cable is
y = - 0.005(x - 45)² + 16.125
A graph of the solution verifies that the solution is correct.