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Mademuasel [1]
2 years ago
9

A restaurant offers 10 appetizers and 12 main courses. In how many ways can a person order a​ two-course meal? Use the Fundament

al Counting Principle with two groups of items.
Mathematics
1 answer:
aliya0001 [1]2 years ago
5 0

Answer:

80 ways

Step-by-step explanation:

The first thing is to apply the Fundamental Counting Principle with two groups of items, in this case the multiplication rule is used.

In this case, what we should do is that, for every 1 main course there are 8 appetizers, but there are 10 main courses therefore:

8 * 10 = 80

There are 80 ways to combine appetizers and the main course

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35 points!!!I need help on these questions it’s a part of a common assessment.
pashok25 [27]

Answer:

1) 4

2) y <= 9

3) NO SOLUTION

4) 24

5) 5 cards

Step-by-step explanation:

1st pic) 2x + 7 = 15

2x + 7 = 15

<u>       -7    -7</u>

2x = 8

divide by 2 to isolate x

<u>x = 4</u>

2nd pic) y - 3 <= 6

y - 3 <= 6

<u>   +3      +3</u>

<u>y <= 9</u>

<u />

3rd pic) x + 0.5 = x + 6

x + 0.5 = x + 6

<u>    -0.5         -0.5</u>

x = x + 5.5

NO SOLUTION

4th pic)  5m + 10 <= 130

5m + 10 <= 130

<u>       -10        -10</u>

5m <= 120

divide by 5 to isolate the m

<u>m <= 24</u>

5th pic) 2c - 8 > 0

2c - 8 > 0

<u>      +8   +8</u>

2c > 8

divide by 2 to isolate the c

<u>c > 4</u>

4 0
3 years ago
Read 2 more answers
Help please is for Algebra
Sauron [17]

Answer:

=\frac{8a+6}{\left(a-3\right)\left(a+3\right)}

the third option is your answer

Step-by-step explanation:

\frac{5}{a-3}+\frac{3}{a+3}\\\mathrm{Least\:Common\:Multiplier\:of\:}a-3,\:a+3:\quad \left(a-3\right)\left(a+3\right)\\\mathrm{Adjust\:Fractions\:based\:on\:the\:LCM}\\=\frac{5\left(a+3\right)}{\left(a-3\right)\left(a+3\right)}+\frac{3\left(a-3\right)}{\left(a+3\right)\left(a-3\right)}\\\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}\\=\frac{5\left(a+3\right)+3\left(a-3\right)}{\left(a-3\right)\left(a+3\right)}

\mathrm{Expand}\:5\left(a+3\right)+3\left(a-3\right):\quad 8a+6\\=\frac{8a+6}{\left(a-3\right)\left(a+3\right)}

6 0
2 years ago
Uncle Jason is twice as old as his nephew Ralph, and 10 years ago he was three times as old as Ralph. Find their present ages.
cluponka [151]

Answer:

40 and 20

Step-by-step explanation:

  • Uncle Jason's age = x
  • Nephew Ralph age = y
<h3>Solution</h3>

<u>Age now:</u>

  • x= 2y

<u>Age 10 years ago: </u>

  • x- 10 = 3(y-10)

<u>Substituting x in the second equation:</u>

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<u>Then, finding x:</u>

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  • x= 2*20
  • x= 40

<u>Answer:</u> Uncle Jason is 40 and nephew Ralph is 20 years old

7 0
2 years ago
What is the horizontal asymptote of the function? f(x)=3x2^x+4
torisob [31]

The vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors

5 0
2 years ago
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melomori [17]
The answer is C,
120x + 60y \geqslant 600
We use the 'more than or equal sign' as the consultant needs to make at least $600 - so exactly $600 or more than $600
4 0
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