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larisa [96]
3 years ago
12

The local diner offers a meal combination consisting of an appetizer, a soup, a main course, and a dessert. There are three appe

tizers, five soups, three main courses, and three desserts. Your diet restricts you to choosing between a dessert and an appetizer. (You cannot have both.) Given this restriction, how many three-course meals are possible
Mathematics
1 answer:
ololo11 [35]3 years ago
3 0

Answer:

number of ways to chose a three course meal is 90

Step-by-step explanation:

Given the data in the question;

Number of appetizer = 3

Number of soups = 5

Number of main courses = 3

Number of desserts = 3

Now,

Number of ways that appetizer can be chosen = ³C₁ = 3!/(1!(3-1)!) = 3

Number of ways that soups can be chosen = ⁵C₁ = 5!/(1!(5-1)!) = 5

Number of ways that main courses can be chosen = ³C₁ = 3!/(1!(3-1)!) = 3

Number of ways that desserts can be chosen = ³C₁ = 3!/(1!(3-1)!) = 3

So,

First we chose an appetizer, soup and main course.

Number of ways will be;

⇒ 3 × 5 × 3 = 45

Next,  we chose a dessert, a soup & a main course.

Number of ways will be;

⇒ 3 × 5 × 3 = 45

Total number of ways to chose a three course meal

⇒ 45 + 45 = 90 ways

Therefore, number of ways to chose a three course meal is 90

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Elimination_4x_2y=14<br> _10x+7y=_24
Gnesinka [82]

Answer:

The answer is x=-\frac{50}{8}  and y=\frac{11}{2}.

Step-by-step explanation:

Given:

-4x-2y=14

-10x+7y=-24

Now, to solve it by elimination:

-4x-2y=14   ......(1)

-10x+7y=-24 ......(2)

So, we multiply the equation (1) by 7 we get:

-28x-14y=98

And, we multiply the equation (2) by 2 we get:

-20x+14y=-48

Now, adding both the new equations:

-28x-14y+(-20x+14y)=98+(-48)

-28x-14y-20x+14y=98-48

-28x-20x-14y+14y=50

-8x=50

<em>Dividing both the sides by -8 we get:</em>

x=-\frac{50}{8}

Now, putting the value of x in equation (1):

-4x-2y=14

-4(-\frac{50}{8})-2y=14

\frac{200}{8} -2y=14

25-2y=14

<em>Subtracting both sides by 25 we get:</em>

-2y=-11

<em>Dividing both sides by -2 we get:</em>

y=\frac{11}{2}

Therefore, the answer is x=-\frac{50}{8}  and y=\frac{11}{2}.

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B4%7D%20%20-%206%20%7Bx%7D%5E%7B3%7D%20%20%2B%2022%20%7Bx%7D%5E%7B2%7D%20%20-%2
alexira [117]

Answer:

x = 2, 1 + 3i, 1 − 3i

Step-by-step explanation:

Find the Roots (Zeros)

x^4 − 6x^3 + 22x^2 − 48x + 40

Set x^4 − 6x^3 + 22x^2 − 48x + 40 equal to 0. x^4 − 6x^3 + 22x^2 − 48x + 40 = 0

Solve for x.

Factor the left side of the equation.

Factor x^4 − 6x^3 + 22x^2 − 48x + 40 using the rational roots test.

(x − 2) (x^3 − 4x^2 + 14x − 20) = 0

 Factor x^3 − 4x^2 + 14x − 20 using the rational roots test.

(x − 2) (x − 2) (x2 − 2x + 10) = 0

 Combine like factors.

(x − 2)2 (x^2 − 2x + 10) = 0

If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.

(x − 2)^2 = 0

x^2 − 2x + 10 = 0

 Set (x − 2)^2 equal to 0 and solve for x.

Set (x − 2)^2 equal to 0.

 (x − 2)^2 = 0

Solve (x − 2)^2 = 0 for x.

x = 2

 Set x^2 − 2x + 10 equal to 0 and solve for x.

Set x^2 − 2x + 10 equal to 0. x^2 − 2x + 10 = 0

Solve x^2 − 2x + 10 = 0 for x.

Use the quadratic formula to find the solutions.

−b ± (√b^2 − 4 (ac) )/2a

Substitute the values a = 1, b = −2, and c = 10 into the quadratic formula and solve for x.

2 ± (√(−2)^2 − 4 ⋅ (1 ⋅ 10))/2 ⋅ 1

Simplify.

Simplify the numerator.

  x =    2 ± 6i/ 2.1

Multiply 2 by 1

 x =  2 ± 6i/2⋅1

 Simplify

  2 ± 6i/2  

   x = 1 ± 3i

The final answer is the combination of both solutions.

x = 1 + 3i, 1 − 3i

The final solution is all the values that make (x − 2)2 (x2 − 2x + 10) = 0 true.

x = 2, 1 + 3i, 1 − 3i

3 0
3 years ago
I do not understand
Pie

Answer:

2500 m^3/1 m^2

Step-by-step explanation:

Going from cm to m is just dividing by 100. So 250,000 cm^3/100= 2,500 m^3

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3 0
2 years ago
X÷-1.2= -0.3 i need help with this asap pls and thank you ​
Aleks [24]

0.3×1.2

=0.36

−0.3,1.2→Negative

−0.36

3 0
2 years ago
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