Answer:
Correct answer is:
Step-by-step explanation:
Given that Number of bracelets with yellow beads is represented by 
Each bracelet with yellow beads is sold for $5.
Total money raised by bracelets with yellow beads = Number of bracelets sold 
Money raised by sale of one such bracelet = 
Also Given that Number of bracelets with Orange beads is represented by 
Each bracelet with orange beads is sold for $6.
Total money raised by bracelets with orange beads = Number of bracelets sold Money raised by sale of one such bracelet = 
Given that total money raised by sale of both type of bracelets is $660.
so, the first equation becomes:
It is also given that "<em>The number of bracelets with yellow beads that Sierra sold is 8 more than twice the number of bracelets with orange beads</em>"
So, by equation (1) and (2), the system of equations is:
Answer:
Step-by-step explanation:
Yes, she can use this inequality and it does matter since the number of cars that the inequality provides will need to be equal to or more than that number in order for all the students to be able to go. Therefore, if we apply the inequality it would give us the minimum number of cars needed (n) like so
12 + 3n > 28 ... subtract 12 on both sides
3n > 16 ... divide both sides by 3
n > 5 1/3
Since there can't be 1/3 of a car and the number of cars needed must be higher than 5 1/3 then we would need a total of 6 cars to take all of the children.
D
Step-by-step explanation:
The number of guests is the independent variable and should be between 1 and 50. We can use a scale of 10's
The cost is the dependent variable and the minimum cost is 100 and should rise from there
The scale would be from 100 's
Answer:
C = (2,2)
Step-by-step explanation:
B = (10 ; 2)
M = (6 ; 2)
C = (x ; y )
|___________|___________|
B (10;2) M (6;2) C ( x; y)
So:
dBM = dMC
√[(2-2)^2 + (6-10)^2] = √[(y-2)^2 + (x - 6)^2]
(2-2)^2 - (6-10)^2 = (y-2)^2 + (x - 6)^2
0 + (-4)^2 = (y-2)^2 + (x - 6)^2
16 = (y-2)^2 + (x - 6)^2
16 - (x - 6)^2 = (y-2)^2
Also:
2*dBM = dBC
2*√[(2-2)^2 + (6-10)^2] = √[(y-2)^2 + (x - 10)^2]
4*[(0)^2 + (-4)^2] = (y-2)^2 + (x - 10)^2
4*(16) = (y-2)^2 + (x - 10)^2
64 = (y-2)^2 + (x - 10)^2
64 = 16 - (x - 6)^2 + (x - 10)^2
48 = (x - 10)^2 - (x - 6)^2
48 = x^2 - 20*x + 100 - x^2 + 12*x - 36
48 = - 20*x + 100 + 12*x - 36
8*x = 16
x = 2
Thus:
16 - (x - 6)^2 = (y-2)^2
16 - (2 - 6)^2 = (y-2)^2
16 - (-4)^2 = (y-2)^2
16 - 16 = (y-2)^2
0 = (y-2)^2
0 = y - 2
2 = y
⇒ C = (2,2)
