Answer:
D.
Step-by-step explanation:
Momentum(p) is Mass(m) times Velocity(v) or p = mv
The results of the calculations are attached. 0.4 kg at 2 m/s has the most momentum: 0.8 kg*m/s
Tom is the deli manager at a grocery store. He needs to schedule employees to staff the deli department at least 260 person-hour
1 answer:
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Answer:
System of inequalities:
,
1) Maximum number of sundaes possible = 9
2) Maximum number of milkshakes possible = 6
3) Combination that uses the most of both Ice-cream and Strawberries = 7 scoop of ice-cream and 1 scoop of strawberries.
Step-by-step explanation:
Given : A sundae requires 3 ice-cream scoops and 4 strawberries, and a milkshake requires 2 ice-cream scoops and 6 strawberries.
Ramses wants to make sundaes and milkshakes with at most 25 ice-cream scoops and 37 strawberries.
Let x denote the number of sundaes he makes and y the number of milkshakes he makes.
First we represent in tabular form,
Sundae(x) Milkshake(y) Total
Ice-cream 3 2 3x+2y
Strawberries 4 6 4x+6y
→System of inequalities:
Sundaes and milkshake with at most 25 ice-cream scoops=
Sundaes and milkshakes with at most 37 strawberries =
→ Plotting the equations in the graph (figure attached)
1) Maximum number of sundaes possible:
Maximum no. of sundaes possible when y=0
From the graph y=0 at x=9.25
Therefore, Maximum number of sundaes possible is 9
2) Maximum number of milkshakes possible:
Maximum no. of milkshakes possible when x=0
From the graph x=0 at y= 6.167
Therefore, Maximum number of milkshakes possible is 6
3) Combination that uses the most of both Ice-cream and Strawberries:
Combination of both is possible there is a intersection of both the equation
From the graph intersection point is x=7.6 and y=1.1
Therefore, Combination that uses the most of both Ice-cream and Strawberries = 7 scoop of ice-cream and 1 scoop of strawberries.
Answer:
Option A is correct (17,11).
Step-by-step explanation:
6x - 9y = 3
3x - 4y =7
it can be represented in matrix form as
A=
X=
B=
i.e, AX=B
or X= A⁻¹ B
A⁻¹ = 1/|A| * Adj A
determinant of A = |A|= (6*-4) - (-9*3)
= (-24)-(-27)
= (-24) + 27 = 3
so, |A| = 3
Adj A=
A⁻¹ = /3
A⁻¹ =
X= A⁻¹ B
X=
X=
X=
X=
x= 17, y= 11
solution set= (17,11).