We are going to make simultaneous equations.
x will be our $3 ice cream and y will be our $5 ice cream
Equation1 ---- x + y = 50 (the sum of all the ice creams they sell)
Equation 2 ---- 3x + 5y = 180 Sum of all the $3 and $5 ice creams is $180
Since we can't solve for both variables we will put one of the variables in terms of the other.
Take x+y=50 and subtract y from both sides. (I could have done subtracted x - it did not matter). Now we have x= ₋ y +50 (negative y +50)
Now I am going to take equation 2 and replace the x with -y +50
3 (-y +50) + 5y = 180
Now I will use the distributive law on the 3 and what's in the parentheses:
-3y + 150 + 5y = 180
Now I will combine like terms (the -3y and the 5y)
2y + 150 = 180
Now subtract 150 from both sides of the equation
2y = 30
Divide both sides by 2
and get y= 15 They sold 15 ice creams that cost $5 each
Since equation 1 is x+y=50 we can replace y with 15
x + 15 = 50 Now subtract 15 from both sides x = 35
Since x represents the $3 ice creams, they sold 35 of those.
Check:
35 X 3 = $105
15 x 5 = + <u>75
</u> $180
7/10 8/11 please tell me if this is helpful, I know 7/10 is right but idk about 8/11
Answer:
0
Step-by-step explanation:
Let's define 3 areas:
- S = area of semicircle with radius 6 in (diameter AB)
- T = area of quarter circle with radius 6√2 in (radius AC)
- U = area of triangle ABC (side lengths 6√2)
The white space between the "moon" and the triangle has area ...
white = T - U
Then the area of the "moon" shape is ...
moon = S -white = S -(T -U) = S -T +U
The area we're asked to find is ...
moon - triangle = (S -T +U) -U = S -T
__
The formula for the area of a circle of radius r is ...
A = πr²
So, ...
S = (1/2)π(6 in)² = 18π in²
and
T = (1/4)π(6√2 in)² = 18π in²
The difference in areas is S -T = (18π in²) -(18π in²) = 0.
There is no difference between the areas of the "moon" and the triangle.
Use desmos
It’s a website
that’ll give you the answer
Answer:
B. (-1,7)
C. (0,5)
D. (2,3)
Step-by-step explanation:
We solve this question looking at the graphic.
A. (-7, 1)
When 
, and thus (7,-1) is not solution of the equation.
B. (-1,7)
When 
, and thus (-1,7) is a solution of the equation.
C. (0,5)
When 
, and thus (0,5) is a solution of the equation.
D. (2,3)
When 
, and thus (2,3) is a solution of the equation.
E. (3, 2)
When 
, and thus (3,2) is not part of the solution
F. (5,0)
When 
, and thus (5,0) is not part of the solution
