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Kitty [74]
3 years ago
9

The menu at jestine's restaurant has side dishes and main dishes. The dishes are rice potatoes and vegetables. The main dishes a

re chicken, fish and beef . If you can order one side dish and one main dish , how many possible combinations are there
A. 6
B. 9
C. 7
D. 12
Mathematics
1 answer:
Dvinal [7]3 years ago
8 0

The given side dishes are rice potatoes and vegetables = 2

The given main dishes are chicken, fish and beef = 3

As there are 2 side dishes. With rice potatoes we can order any of the three main dish, so it is 3 combination. Similarly with vegetables also, we can order any of the three main dish, so it is 3 combination.

So, combined combination is 3+3 = 6

Hence, the correct answer is option A.


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If f(x)=3^2 and g(x)=4x^3+1 , what is the degree of (f*g)(x)?
N76 [4]
I'm assuming you meant to say f(x) = 3x^2. If that assumption is correct, then the degree is found by multiplying the leading terms from both f(x) and g(x). The leading terms are 3x^2 and 4x^3

3x^2*4x^3 = (3*4)*(x^2*x^3) = 12x^(2+3) = 12x^5

The exponent of that result is 5, so the degree of (f*g)(x) is 5

Answer: 5
5 0
3 years ago
A map uses a scale of 3/4 in = 80 miles. If two buildings are 60 miles apart in real life, how far apart would they be drawn on
Alla [95]

Given:

The scale of a map is \dfrac{3}{4}\ in=80\ miles.

The distance between two buildings is 60 miles.

To find:

The distance between the given building on the map.

Solution:

We have,

\dfrac{3}{4}\ in=80\ miles

It means, 80 miles in real life = \dfrac{3}{4}\ in on the map.

1 mile in real life = \dfrac{3}{4\times 80}\ in on the map.

60 mile in real life = \dfrac{3\times 60}{4\times 80}\ in on the map.

                              = \dfrac{180}{320}\ in on the map.

                              = \dfrac{9}{16}\ in on the map.

Therefore, the buildings are \dfrac{9}{16}\ in apart on the map.

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