Answer:
2 barbies are left over
Step-by-step explanation:
5(5)=25
27-25=2
Leaving 2 barbies.
Answer:
C
Step-by-step explanation:
Answer: c
Step-by-step explanation:
I just took the test and got it correct
Answer:
m=1
Step-by-step explanation:
The plots (-4,3) and (2,1) have a slope of -1. Perpendicular lines would mirror that, making it a positive 1
m∠3+m∠8=180
∠4-∠7
∠1=∠8
Answer: a) 2 miles
b) 4 miles
Step-by-step explanation:
There are two right angle triangles formed in the rectangle.
Taking 30 degrees as the reference angle, the length of the side walk, h represents the hypotenuse of the right angle triangle.
The width, w of the park represents the opposite side of the right angle triangle.
The length of the park represents the adjacent side of the right angle triangle.
a) to determine the width of the park w, we would apply
the tangent trigonometric ratio.
Tan θ, = opposite side/adjacent side. Therefore,
Tan 30 = w/2√3
1/√3 = w/2√3
w = 1/√3 × 2√3
w = 2
b) to determine the the length of the side walk h, we would apply
the Cosine trigonometric ratio.
Cos θ, = adjacent side/hypotenuse. Therefore,
Cos 30 = 2√3/h
√3/2 = 2√3/h
h = 2√3 × 2/√3
h = 4
Answer:
area = 1500× 750 =
Step-by-step explanation:
we know area of rectangle
for length = l m
and width = b m
and perimeter
but one side length measures is not required because of the river so
He does not use the fence along the side of the river
so we use this formula
Perimeter = P = L + 2 b
Perimeter is 3000 m
so area will be
it is a quadratic function whose max or min will
occur at the average of the Solutions.
on Solving (3000 - 2b)b = 0
3000 - 2b = 0 or b=0
2b =3000
or
The average of the values are
so for max area we use b=
The Length is then L=3000 - 2(750) = 3000 - 1500 = 1500
for max area
length = 1500 m
bredth = 750 m
area = 1500× 750 =
The largest area that can be enclosed by Farmer Ed with 3000 meters of fencing along a river (with only three sides fenced) equals 1,125,000 square meters by using principles of mathematical optimization.
In this question, Farmer Ed wants to maximize the area of a rectangle with only three sides fenced, since one side borders on a river. We can use the principles of optimization in mathematics to solve this problem.
With 3000 meters of fencing for three sides, if we denote one side perpendicular to the river as X and the side parallel to the river (which forms the base of the rectangle) as Y, then, the perimeter would be Y+2X which is equal to 3000 meters. So, Y = 3000-2X.
The area A of a rectangle is length times width, or, in this case, A = XY. Substituting Y from the equation above: A = X(3000-2X) = 3000X - 2X^2. To maximize this area, we need to find values of X for which this equation has its maximum value.
The maximum or minimum of a function can be found at points where its derivative is zero. So, we take the derivative of A with respect to X, set it equal to zero, and solve for X.
The derivative, dA/dX is 3000 - 4X. Setting this equal to 0 gives X = 3000/4 = 750. So, the maximum area that Farmer Ed can enclose is when X is 750, and Y is 3000 - 2X = 1500, so the maximum area is 750 * 1500 = 1,125,000 square meters.
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