1. Make a factor tree( always helps for me).
2. Identify the common numbers that make it.
3. 36 and 54 have factors of 2 and 3s in common.
4. Multiply the common factors to find the GCF.
I hope this helped!:)
(a) Find a function that gives the area A(x) of the playground (in square meters) in terms of x.
(b) What side length x gives the maximum area that the playground can have?
(c) What is the maximum area that the playground can have?
Answer:
Part a)
Part b) The side length x that give the maximum area is 120 meters
Part c) The maximum area is 14,400 square meters
Step-by-step explanation:
The picture of the question in the attached figure
Part a) Find a function that gives the area A(x) of the playground (in square meters) in terms of x
we know that
The perimeter of the rectangular playground is given by
we have
substitute
solve for W
Find the area of the rectangular playground
The area is given by
we have
substitute
Convert to function notation
Part b) What side length x gives the maximum area that the playground can have?
we have
This function represent a vertical parabola open downward (the leading coefficient is negative)
The vertex represent a maximum
The x-coordinate of the vertex represent the length that give the maximum area that the playground can have
Convert the quadratic equation into vertex form
Factor -1
Complete the square
The vertex is the point (120,14,400)
therefore
The side length x that give the maximum area is 120 meters
Part c) What is the maximum area that the playground can have?
we know that
The y-coordinate of the vertex represent the maximum area
The vertex is the point (120,14,400) -----> see part b)
therefore
The maximum area is 14,400 square meters
Verify
The playground is a square
The width of the playground is 120 meters, the side length that gives the maximum area is 120 meters, and the maximum area the playground can have is 14400 square meters.
(a) Let's assume the width of the rectangle is x meters. Since the playground is rectangular and has two equal sides, the length will also be x meters. The perimeter of the rectangle, which is also the amount of fencing needed, is given as 480 meters. This can be expressed as: 2(length + width) = 480. Using this equation, we can solve for the width: 2(x + x) = 480 ⇒ 4x = 480 ⇒ x = 480/4 = 120. Therefore, the width of the playground is 120 meters.
(b) To find the side length that gives the maximum area, we can use calculus. The area function is A(x) = x * x = x^2. To find the maximum of this function, we can take the derivative and set it equal to zero: dA/dx = 2x = 0 ⇒ x = 0. So, x = 0 is a critical point, but since we are dealing with a physical situation where the length cannot be zero, we disregard this critical point. Thus, x = 120 is the value that gives the maximum area.
(c) Now that we know the side length, we can calculate the maximum area. Plugging in x = 120 into the area function, we find: A(120) = 120 * 120 = 14400 square meters. Therefore, the maximum area the playground can have is 14400 square meters.
#SPJ12
The length of the diagonals of the rectangle are AB = 30 and BD = 30
A 4-sided flat shape with straight sides where all interior angles are right angles (90°). Also, the opposite sides are parallel and of equal length.
Given that, ABCD is a rectangle and its diagonals are AC= 3(x-2) and BD=x+18
According to the property of a rectangle,
Diagonals of a rectangle are equal.
Therefore, AC = BD
3(x-2) = x+18
3x-6 = x+18
2x = 24
x = 12
AC = 3(12-10) = 30
BD = 12+18 = 30
Hence, the diagonals of the rectangle are AB = 30 and BD = 30
For more references on rectangles, click;
#SPJ2
The diagonals of a rectangle are congruent. That means that their lengths are equal. One diagonal is AC, and the other diagonal is BD. AC must equal BD. We set their lengths equal and solve for x.
3(x - 2) = x + 18
Distribute the 3 on the left side.
3x - 6 = x + 18
Subtract x from both sides; add 6 to both sides.
2x = 24
Divide both sides by 2.
x = 12
Now that we know x = 12, we replace x with 12 in BD = x + 18 to find the length of BD.
BD = x + 18 = 12 + 18 = 30
Since the diagonals are congruent, the length of AC is also 30.
Answer: AC = 30; BD = 30
Ratio is the relationship between the number of same unit to verify the how bigger or smaller is a number to the another when we compare two quantities to each we write them in the form of the ration. The ratio of local calls to long distance calls made by Jillian in that month is 9/5. Thus the option C is the correct option.
Total local phone calls made by the Jillian is 36.
Total long distance phone calls made by the Jillian is 20.
To get the value of ratio we need to know about the term "ratio".
Ratio is the relationship between the number of same unit to verify the how bigger or smaller is a number to the another when we compare two quantities to each we write them in the form of the ratio.
Now to find out the ratio of local calls to long distance calls made by Jillian in that month we need to divide the number of local calls to the number of long distance calls.
Suppose r be the ratio of local calls to long distance calls made by Jillian in that month. Hence.
Hence the ratio of local calls to long distance calls made by Jillian in that month is 9/5. Thus the option C is the correct option.
Learn more about the ratio here;
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Explanation:
We'll be using the n C r combination function. To make the notation a bit easier to deal with, I will write "C(n,r)" instead of "n C r".
The formula is
C(n,r) = (n!)/(r!*(n-r)!)
where the exclamation marks represent factorials.
A factorial is where you start with a positive integer, and count down to 1 multiplying all along the way.
Examples:
5! = 5*4*3*2*1
8! = 8*7*6*5*4*3*2*1 = 120
Note how the string "5*4*3*2*1" is in both 5! and 8!
We can say 8! = 8*7*6*5!
Because we can replace the "5!" at the end with "5*4*3*2*1" later if we wanted. This strategy is used to help find a shortcut to simplification.
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We have n = 9 appetizers and r = 6 items we can select from this pool.
C(n,r) = (n!)/(r!*(n-r)!)
C(9,6) = (9!)/(6!*(9-6)!)
C(9,6) = (9!)/(6!*3!)
C(9,6) = (9*8*7*6!)/(6!*3*2*1)
C(9,6) = (9*8*7)/(3*2*1) .... the "6!" terms canceled out
C(9,6) = 504/6
C(9,6) = 84
There are 84 ways to choose six appetizers from the pool of nine available
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Repeat those steps for the main courses. Use n = 8 and r = 5 this time.
C(n,r) = (n!)/(r!(n-r)!)
C(8,5) = (8!)/(5!*(8-5)!)
C(8,5) = (8!)/(5!*3!)
C(8,5) = (8*7*6*5!)/(5!*3*2*1)
C(8,5) = (8*7*6)/(3*2*1)
C(8,5) = (336)/(6)
C(8,5) = 56
There are 56 ways to choose five main course meals from the pool of eight available
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Then do the same for the desserts. Use n = 3 and r = 2.
C(n,r) = (n!)/(r!(n-r)!)
C(3,2) = (3!)/(2!*(3-2)!)
C(3,2) = (3!)/(2!*1!)
C(3,2) = (3*2*1)/(2*1*1)
C(3,2) = 6/3
C(3,2) = 3
There are 3 ways to choose two desserts from the pool of three available
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The last step is to multiply all these results:
84*56*3 = 14112
This is the number of ways to select all of the items given the restrictions listed. The order does not matter.