The total length of red and blue ribbons is 10.2 m.
Given that, the length of the red ribbon =2.4 m and the length of the blue ribbon =7.8 m.
<h3>How to add two decimals?</h3>
Adding decimals is similar to the usual addition of whole numbers. We know that a decimal number is a number with a whole number part and a fractional part which is separated by a decimal point.
Now, the length of the red ribbon + the length of the blue ribbon=2.4+7.8
=10.2 m
Therefore, the total length of red and blue ribbons is 10.2 m.
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Answer:
Step-by-step explanation:
a² - b² = (a +b)(a - b)
9x² - 4 = (3x)² - 2²
= (3x + 2 )(3x - 2)
(25-y) ^2-y^2=25
y=12
x=25-12
x=13
(13,12)
Answer:
y = -2x + 6
Step-by-step explanation:
y=mx + c
m is the gradient and c is the y intercept
m = (y 2 - y 1) / (x 2 - x 1) = (2-4) / ( 2-1) = -2
y= -2x + c
To find c, just sub one of the coordinates into the eqn:
By using the coordinates (1,4)
4 = -2(1) + c
c = 6
Therefore, the equation is y = -2x + 6
Answer:
1. Objective function is a maximum at (16,0), Z = 4x+4y = 4(16) + 4(0) = 64
2. Objective function is at a maximum at (5,3), Z=3x+2y=3(5)+2(3)=21
Step-by-step explanation:
1. Maximize: P = 4x +4y
Subject to: 2x + y ≤ 20
x + 2y ≤ 16
x, y ≥ 0
Plot the constraints and the objective function Z, or P=4x+4y)
Push the objective function to the limit permitted by the feasible region to find the maximum.
Answer: Objective function is a maximum at (16,0),
Z = 4x+4y = 4(16) + 4(0) = 64
2. Maximize P = 3x + 2y
Subject to x + y ≤ 8
2x + y ≤ 13
x ≥ 0, y ≥ 0
Plot the constraints and the objective function Z, or P=3x+2y.
Push the objective function to the limit in the increase + direction permitted by the feasible region to find the maximum intersection.
Answer: Objective function is at a maximum at (5,3),
Z = 3x+2y = 3(5)+2(3) = 21