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lisabon 2012 [21]

3 years ago

6

A menu offers a choice of 22 salads, 77 main dishes, and 55 desserts. How many different meals consisting of one salad, one m

ain dish, and one dessert are possible?

1 answer:

weqwewe [10]3 years ago

8 0

Answer:

70 different meals

Step-by-step explanation:

A menu offers a choice of 2 salads, 7 main dishes, and 5 desserts

There are 3 types of choices are given

each one have different choices. we need to select one salad, one main dish and one dessert

To find the number of different meals we multiply the different choices

$2 \cdot 7 \cdot 5=70$

so 70 different meals

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While driving from Chicago, Brent had a visibility problem due to the fog. After 1 1/2

liq [111]

1) Let x be the original speed that he was driving;

x mph ---> 1.5 hours
x+44 mph --> 2.5 hours

1.5x + 2.5(x+44)=206
1.5x+2.5x+110=206
4x=206-110
4x=96
x=24
Therefore, he was travelling at 24 mph in the fog.

2) Let x be the speed of 1 snail,
The other snail travels at the speed of x+1 cm/min

23x+23(x+1)=391 <-- solve for x
23x+23x+23=391
46x=368
x=368/46
x=8

Therefore one of the snail travels at 8 cm/min while the other goes at 9 cm/min

7) Let x be one crew
Then the other crew would be working at x+0.5

3.2x+3.2(x0.5)=8 <-- solve for x
3.2x+1.6x=8
4.8x=8
x=8/4.8
x=1.6666 or 1.7 miles

Therefore, one of the crew does 1.7 miles per day and the other crew does (0.5*1.7)=0.85 miles

Hope I helped :)

5 0

3 years ago

How do you translate the equation 4x-5=3(x+2) into a verbal sentence

elena55 [62]

Four times the quantity of x minus five is equal to three times the quantity of x plus two

3 0

2 years ago

Find the value of c .<br> 5/2 = c/22<br> c = ___.

sleet_krkn [62]

Answer:

c = 55

Step-by-step explanation:

8 0

3 years ago

For each of the following vector fields

olga nikolaevna [1]

(A)

$\dfrac{\partial f}{\partial x}=-16x+2y$

$\implies f(x,y)=-8x^2+2xy+g(y)$

$\implies\dfrac{\partial f}{\partial y}=2x+\dfrac{\mathrm dg}{\mathrm dy}=2x+10y$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=10y$

$\implies g(y)=5y^2+C$

$\implies f(x,y)=\boxed{-8x^2+2xy+5y^2+C}$

(B)

$\dfrac{\partial f}{\partial x}=-8y$

$\implies f(x,y)=-8xy+g(y)$

$\implies\dfrac{\partial f}{\partial y}=-8x+\dfrac{\mathrm dg}{\mathrm dy}=-7x$

$\implies \dfrac{\mathrm dg}{\mathrm dy}=x$

But we assume $g(y)$ is a function of $y$ alone, so there is not potential function here.

(C)

$\dfrac{\partial f}{\partial x}=-8\sin y$

$\implies f(x,y)=-8x\sin y+g(x,y)$

$\implies\dfrac{\partial f}{\partial y}=-8x\cos y+\dfrac{\mathrm dg}{\mathrm dy}=4y-8x\cos y$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=4y$

$\implies g(y)=2y^2+C$

$\implies f(x,y)=\boxed{-8x\sin y+2y^2+C}$

For (A) and (C), we have $f(0,0)=0$ , which makes $C=0$ for both.

4 0

2 years ago

Can someone solve this question?

iren2701 [21]

It’s 26 bc it’s says it right there

5 0

2 years ago