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

Answer 1

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

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

Answer 2

Answer: [(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

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Which is equivalent to (9y2 – 4x)(9y2 + 4x), and what type of special product is it?81y4 – 16x2, a perfect square trinomial
81y4 – 16x2, the difference of squares
81y4 – 72xy2 – 16x2, a perfect square trinomial
81y4 – 72xy2 – 16x216, the difference of squares

Answers

Consider the expression $(9y^2- 4x)(9y^2 + 4x) .$ You can see that in both brackets are the same terms $9y^2$ and $4x$ , but signs are different. So you can use formula for the difference of squares:

$a^2-b^2=(a-b)(a+b).$

According to this formula,

$(9y^2-4x)(9y^2+4x)=(9y^2)^2-(4x)^2=9^2\cdot (y^2)^2-4^2\cdot x^2=81y^4-16x^2.$

Answer: correct choice is B.

Plz help will mark you as brainliest(the one who answers the most gets brainliest) u don't have to answer all of them u can just answer 2 or 3

Answers

Answer:

41. s = semiperimeter

s = (a + b + c) /2

s = (120+170+250) /2

s = 270m

d is the correct option.

42. b is the correct option.

43. P = a + b + c

P = 120+170+250

P = 540m

a is the correct option.

44. A = √s((s-a)(s-b)(s-c))

=√270((270-120)(270-170)(270-250))

=9000m²

d is the correct option.

45. length of wire=P - space for gate

= 540 - 3

=537m

b is the correct option.

5/12 - 2/5 in fraction form

Answers

5/12-2/5=25/60-24/60
1/60

the answer is 1/60

25/60-24/60
So, the correct answer is 1/60

The table represents a function.
What is f(5)?
A. -8
B. -1
C. 1
D. 8

Answers

Answer:

A. -8

Step-by-step explanation:

We can look at the table and figure out how we get f(5)

really its just the x value pugged into the x value for the function

after that we look on the table, f(5) is -8 because at the bottum of the table it shows 5 and -8

High Hopes^^

Barry-

The answer for me is the answer -8

To the nearest hundredth of a second, how much time does it take a car to travel 50 feet from a stop at a constant acceleration of 3. 2 ft/s²? Use d=1/2at², where d represents distance, a represents acceleration, and t represents time in seconds. (pic above)

Answers

$d=(1)/(2)at^2\n\nWe\ have:\n\nd=50\ ft,\ a=3.2(ft)/(s^2)\n\nSubstitute:\n\n(1)/(2)\left(3.2(ft)/(s^2)\right)t^2=50\ ft\n\n1.6(ft)/(s^2)t=50\ ft\qquad\text{divide both sides by}\ 1.6(ft)/(s^2)\n\nt^2=50\ ft\cdot(1)/(1.6)(s^2)/(ft)\n\nt^2=(50)/(1.6)s^2\n\nt^2=(500)/(16)s^2\to t=\sqrt{(500)/(16)s^2}\n\nt=(√(500))/(√(16))s\n\nt=(√(100\cdot5))/(4)s\n\nt=(√(100)\cdot\sqrt5)/(4)s\n\nt=(10\sqrt5)/(4)s\approx5.59\ s$

$Answer:\ \boxed{5.59\ s}$

Final answer:

Using the given equation and values, we can calculate that it would take approximately 5.59 seconds for a car to travel 50 feet from a stop with a constant acceleration of 3.2 ft/s².

Explanation:

To solve this physics problem, we'll need to use the equation d=1/2at². Here, 'd' represents distance, 'a' represents acceleration, and 't' represents time. Here, we are given 'd' as 50 feet and 'a' as 3.2 ft/s². Now we just need to isolate 't' and solve for time.

Start with the given equation: d=1/2at².
Substitute the given values: 50=1/2*3.2*t².
Remove the fraction by multiplying all terms by 2: 100=3.2*t².
Divide both sides by 3.2 to isolate t²: t²=31.25.
Take the square root of both sides to solve for 't': t=√31.25.
So, t ≈ 5.59 s (rounded to the nearest hundredth).

To summarize, a car traveling 50 feet from a stop with a constant acceleration of 3.2 ft/s² would take approximately 5.59 seconds. Always remember to isolate the unknown in your equation and solve step-by-step for these types of problems.

Learn more about Constant Acceleration here:

brainly.com/question/37024881

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SAT HelpJack can paint a house in 5 days, and Richard can paint the same house in 7 days. Working together, how long will it take them to finish the job?

How do you solve this?

Answers

Well, their speeds are ( $V_1$ is Jack's speed, and $V_2$ is Richard's.
$V_1 = (1)/(5) houses/day \n V_2 = (1)/(7) houses/day \n V = V_1 + V_2 = (1)/(5)+(1)/(7) = (7+5)/(35) = (12)/(35)$
They, together, can paint 12 houses in 35 days. To get a single house, we only have to calculate $(35)/(12)$ which is very close to 3 (a bit below)

Answer:

In 7*5 = 35 days, Jack can paint 7 houses.

In 5*7 = 35 days, Richard can paint 5 houses.

So in 35 days, the two of them can paint 12 houses. To paint just one house, they'll need 1/12 the time, or 35/12 = 2 11/12 days.

Step-by-step explanation:

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