Problem 3: Let x = price of bag of pretzels Let y = price of box of granola bars
We have Lesley's purchase: 4x+2y=13.50
And Landon's: 1x+5y=17.55
We can use the elimination method. Let's negate Landon's purchase by multiplying by -1. -1x-5y=-17.55
We add this four times to Lesley's purchase to eliminate the x variable.
2y-20y=13.50-70.2
-18y=-56.7
y = $3.15 = Price of box of granola bars
Plug back into Landon's purchase to solve for pretzels.
x+5*3.15=17.55
x+15.75=17.55
x = $1.80 = price of bag of pretzels
Problem 4.
Let w = number of wood bats sold
Let m = number of metal bats sold
From sales information we have: w + m = 23
24w+30m=606
Substitution works well here. Solve for w in the first equation, w = 23 - m, and plug this into the second.
24*(23-m)+30m=606
552-24m+30m=606
6m=54
m=9 = number of metal bats sold
Therefore since w = 23-m, w = 23-9 = 14. 14 wooden bats were sold.
Answer:
a =7.5
Step-by-step explanation:
Since this is a right triangle, we can use the Pythagorean theorem
a^2+ b^2 = c^2 where a and b are the legs and c is the hypotenuse
a^2 + 10 ^2 = 12.5^2
a^2 + 100 =156.25
Subtract 100 from each side
a^2 = 56.25
Take the square root of each side
sqrt(a^2) = sqrt( 56.25)
a =7.5
The answer would be 16 S'mores and the limiting reactant would be the grahams.
(This is assuming that S'mores would need 2 grahams, 1 marshmallow and 1 chocolate piece.)
Limiting reactant would be the reactant that runs out first.
Let's take your problem into account and see what we have:
48 marshallows
32 grahams (2 x 16 per pack)
45 chocolate pieces (5 x 15 pieces per bar)
Since need 2 of the grahams per S'more then the maximum yield of the grahams is 16 S'mores.
The maximum yield of marshmallows is 48.
The maximum yield of chocolate is 45.
Since you cannot make S'mores without the grahams, then you can only make 16 S'mores before the grahams run out.
Answer:
Option 3rd is correct
Step-by-step explanation:
(1,-6),(2,-4),(2,-2),(4,-2),(6,0)
