If a square has an area of 225 square feet what is the length of each side

Answers

Answer 1

Answer: The area of the square: $A=a^2$

where "a" is a length of side.

Therefore:
$A=225\ ft^2\ and\ A=a^2\Rightarrow a^2=225\Rightarrow a=√(225)\Rightarrow \boxed{a=15\ (ft)}$

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18-6and -18+6 is the same thing

Answers

Answer:

Step-by-step explanation:

18-6=12

-18+6=-12

It is the opposite because rather than subtracting from a positive, you're adding to a negative.

Hope this helps :)

={125},={#125}andQ={#125},
finda. ∪ b. ∩ c. d.

Answers

Answer:

h3llo

Step-by-step explanation:

hey the answer is in your heart

Simplify square root parenthesis 1 minus sine theta parenthesis times parenthesis 1 plus sine theta parenthesis.

Answers

Answer:

$cos{\theta}$

Step-by-step explanation:

The given equation will be of the form:

$\sqrt{(1-sin{\theta})(1+sin{\theta})}$

Simplifying the above equation, we get

Using identity, $(a+b)(a-b)=a^2-b^2$

= $\sqrt{1^2-(sin{\theta})^2}$

= $\sqrt{1-sin^2{\theta}}$

= $\sqrt{cos^2{\theta}}$

= $cos{\theta}$

which is the required simplified form of the given equation.

answer is Cos theta
find the attachments for detailed solution

A catering service offers 12 appetizers, 9 main courses, and 7 desserts. A banquet chairperson is to select 8 appetizers, 8 main courses, and 6 desserts for a banquet. In how many ways can this be done?

Answers

$12-appetizers, \ 9- main\ courses, \ 7\ desserts\n \n selection:\ 8\ appetizers,\ 8\ main\ courses, 6\ desserts\n\na=the\ number\ of\ selection\ of\ appetizers:\n \n{12 \choose 8}= (12!)/(8!\cdot (12-8)!) = (12\cdot11\cdot10\cdot9\cdot8!)/(8!\cdot4\cdot3\cdot2) =11\cdot5\cdot9=495\n \nc=the\ number\ of\ selection\ of\ main\ courses:\n \n{9 \choose 8}= (9!)/(8!\cdot (9-8)!) = (9\cdot8!)/(8!\cdot 1) =9$

$d=the\ number\ of\ selection\ of\ desserts:\n \n{7 \choose 6}= (7!)/(6!\cdot (7-6)!) = (7\cdot6!)/(6!\cdot 1) =7\n \nthe\ number\ of\ selection\ sets\ the\ banquet:\n \na\cdot c\cdot d=495\cdot9\cdot7=31185$

[(12! / (8!*4!) ]* [9! / (8!*1!) ] * [ 7!/ (6!*1!)] = ( 12 * 11 * 10 * 9 / 4 * 3 * 2 * 1) * 9 * 7 = 45 * 11 * 9 * 7 = 31185 ways

a pipe cleaner 20cm long. it is bent into a rectangle. use a quadratic model to calculate the dimensions that give the maximum area.

Answers

The dimensions that give the maximum area is 5 cm by 5 cm.

Given:

The perimeter of this rectangle is 20 cm, and formula for perimeter is

P= 2(W+L)

P = 20 cm = 2W + 2L.

Then W + L = 10 cm,

or W = (10 cm) - L.

The area of the rectangle is A = L·W, and is to be maximized.

On substituting the values, we get A = L[ (10 cm) - L ], or A = 10L - L²

Note that this is the equation of a parabola that opens down. With coefficients a = -1, b = 10 and C = 0, we find that the x-coordinate of the vertex (which is the x-coordinate of the maximum as well) is

x = -b / (2a). Subbing 10 for b and -1 for a, we get:

x = -[10] / [2·(-1)] = 10/2, or 5.

This tells us that one dimension of the rectangle is 5 cm.

Since P = 20 cm = 2L + 2W, and if we let L = 5 cm, we get:

20 cm = 2(5 cm) + 2W, or

10 cm = W + 5 cm, or W = 5 cm.

Therefore, choosing L = 5 cm and W = 5 cm results in a square, which in turn leads to the rectangle having the maximum possible area.

Learn more:

brainly.com/question/24639460

Answer:

5 cm by 5 cm

Step-by-step explanation:

The perimeter of this rectangle is 20 cm, and the relevant formula is

P = 20 cm = 2W + 2L. Then W + L = 10 cm, or W = (10 cm) - L.

The area of the rectangle is A = L·W, and is to be maximized. Subbing (10 cm) - L for W, we get A = L[ (10 cm) - L ], or A = 10L - L²

Note that this is the equation of a parabola that opens down. With coefficients a = -1, b = 10 and C = 0, we find that the x-coordinate of the vertex (which is the x-coordinate of the maximum as well) is

x = -b / (2a). Subbing 10 for b and -1 for a, we get:

x = -[10] / [2·(-1)] = 10/2, or 5.

This tells us that one dimension of the rectangle is 5 cm.

Since P = 20 cm = 2L + 2W, and if we let L = 5 cm, we get:

20 cm = 2(5 cm) + 2W, or

10 cm = W + 5 cm, or W = 5 cm.

Thus, choosing L = 5 cm and W = 5 cm results in a square, which in turn leads to the rectangle having the maximum possible area.

If someone could do this it’s be greatly appreciated!

Answers

Answer:

a). A = -2x² + 200x

b). Widths = 40 feet and 60 feet

Step-by-step explanation:

It is given that length of the fencing material = 200 feet

a). Peg wants to cover the vegetable garden from three sides with the given fencing material.

If length of the garden = l

and width of the garden = x

l + x + x = 200

l + 2x = 200

l = (200 - 2x) feet

Therefore, area of the garden = Length × width

A = $(200-2x)* x$

A = -2x² + 200x

b). Foe A = 4800 square feet,

4800 = -2x² + 200x

2x² - 200x + 4800 = 0

x² - 100x + 2400 = 0

x² - 60x - 40x + 2400 = 0

x(x - 60) - 40(x - 60) = 0

(x - 60)(x - 40) = 0

x = 40, 60 feet

Therefore, widths of Peg's garden will be 40 feet and 60 feet.

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