C
Look at how it’s graphed and find your domain and then graph
Mark brainliest please
Answer:
Student ticket: $2
Adult ticket: $5
Step-by-step explanation:
1. Setting up the system of equations
let x be the cost for a child
let y be the cost for an adult
For the monday, that would be:
33x + 42y = 276
For the tuesday, you ahve
46x + 40y = 292
2. Solving
First, simplify each equation
33x + 42y = 276
To get all the numbers smaller divide both sides by 3
11x + 14y = 92
Do the same for the other equation:
46x + 40y = 292
divide both sides by 2
23x + 20y = 146
You want to use elimination, so take the equation 11x + 14y = 92 and multiply both sides by 10
110x + 140y = 920
Take the equation 23x + 20y = 146 and multiply both sides by 7
161x + 140y = 1022
Subtract the two equations:
110x + 140y = 920
- (161x + 140y = 1022)
-51x = -102
divide both sides by -51
x = 2
Plug this into the original equation, 11x + 14y = 92:
11 * 2 + 14y = 92
22 + 14y = 92
subtract 22 from both sides
14y = 70
divide both sides by 14
y = 5
So the cost of a student ticket is $2 and the cost of an adult ticket is $5
<em>Six (6)</em> different pizzas can be made with one vegetable and one meat.
They are:
1). broccoli, pepperoni
2). broccoli, sausage
3). peppers, pepperoni
4). peppers, sausage
5). onions, pepperoni
6). onions, sausage
Estimate = 700 lbs
Real bear = 634 lbs
Difference = (700 - 634) = 66 lbs
The estimate is <em>66 pounds more</em> than the actual weight.
The estimate is 10.4% too high.
The actual bear weighs 9.4% less than the estimate.
The answer would be C. Obtuse Iscosceles
First, lets do all calculations in kg. 1kg=1000 grams
Lets set 1 cat's mass as x.
Lets set the other cat's mass as y.
1) x+y=11
2) y=x+1.5
Lets plug eq 2 into eq 1.
x+x+1.5=11
2x+1.5=11
2x=9.5
x=4.75 kg
If one cat weighs 4.75 kg, then the other must weigh 6.25 kg.