Answer:
Step-by-step explanation:
We have the equations
4x + 3y = 18 where x = the side of the square and y = the side of the triangle
For the areas:
A = x^2 + √3y/2* y/2
A = x^2 + √3y^2/4
From the first equation x = (18 - 3y)/4
So substituting in the area equation:
A = [ (18 - 3y)/4]^2 + √3y^2/4
A = (18 - 3y)^2 / 16 + √3y^2/4
Now for maximum / minimum area the derivative = 0 so we have
A' = 1/16 * 2(18 - 3y) * -3 + 1/4 * 2√3 y = 0
-3/8 (18 - 3y) + √3 y /2 = 0
-27/4 + 9y/8 + √3y /2 = 0
-54 + 9y + 4√3y = 0
y = 54 / 15.93
= 3.39 metres
So x = (18-3(3.39) / 4 = 1.96.
This is a minimum value for x.
So the total length of wire the square for minimum total area is 4 * 1.96
= 7.84 m
There is no maximum area as the equation for the total area is a quadratic with a positive leading coefficient.
The answer is -60. this is because the signs implemented in front of numbers give it a sort pattern .
when there are two additions the last number after the positive sign will give a positive number 1 greater than the initial number for example + 5 + 6. Answer afterward will be + 7 and for negative sign eg: - 7 - 8 in the question will give the answer -8 the same number after a negative (-) sign .
Hope this helps
1) The outcomes for rolling two dice, the sample space, is as follows:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
There are 36 outcomes in the sample space.
2) The ways to roll an odd sum when rolling two dice are:
(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5). There are 18 outcomes in this event.
3) The probability of rolling an odd sum is 18/36 = 1/2 = 0.5