5 eggs weigh 2/3 of a pound. How much does each egg weigh?

If y varies directly with x, the constant of variation is . The equation representing this relationship is . What is the value o

Julli [10]

Answer:

Y~x

Y = k × v

Y = 22

Step-by-step explanation:

Y~ x

Y = k × X

Since the question isn't complete,we are going to assign figures to complete this very question and move on to other things and you can as well solve questions that are similar to this ones.

Now let's assume that the question says that is 6 when x is 3,what is y when x is 11.

To solve this question above,we need to find the constant of variation which will then enable us to know what y is when x is 11

Y~ x

Y = k × V

When y is 6 and x is 3,k will be

6 = 3k

K= 2

Now when k is 2,x is 11,y Will then be

Y = kx

Y = 2 × 11

Y = 22.

With this procedure,one can find any value of x or y when they have the constant of variation.

If you have a figure assigned to x and y is to be found,the constant of variation will help to get y and vice versa

8 0

3 years ago

At one time the price of gasoline was $1.279 a gallon. Then the price

ELEN [110]

Answer:

1.6627 (money wise, $1.66)

Step-by-step explanation:

Since its 130%, you know there was already a whole increase. So first you would do 1.279 + 1.279, which would give you 2.558. Then you need to calculate 30% of 1.279 for the rest of the increase. That gives you 0.3837 Add 2.558 and 0.3837 together, which would give you 2.9417. To get the INCREASE, you would have to subtract the new total from the original price, so 2.9417 - 1.279. This would give you 1.6627.

5 0

2 years ago

Explain why you do not have to do computing to solve 15×0×(13+7)

neonofarm [45]

The answer is 300 if you do the per enthuses first

5 0

3 years ago

Sally lives in City A and works for the department of transportation. She needs to inspect the weigh stations in each of four di

galben [10]

Answer:

She can visit the cities in 192 ways

Step-by-step explanation:

If Sally starts from city A and after going to 4 cities she returns to home,

then that can be in 4! = 24 ways. (permutation of 4 distinct objects)

If between the time she visits cities, she comes home once,

then cities can be visited in,

$4! \times 3_{C_{1}}$ = 72 ways. (permutation of 4 distinct objects and choosing 1 gap among the 3 )

If between the time she visits cities, she comes home twice,

then cities can be visited in,

$4! \times 3_{C_{2}}$ =72 ways. (permutation of 4 distinct objects and finding 2 gaps among the 3)

If between the time she visits cities, she comes home thrice,

then cities can be visited in,

$4! \times 3_{C_{3}}$ = 24 ways. (permutation of 4 distinct objects and choosing 3 gaps among the 3)

Now, she can't come home more than thrice in between the time she visits cities (since there are 4 cities to visit only)

So, the number of ways she can visit cities = (24+ 72 + 72 + 24)

= 192

6 0

3 years ago

How do you solve (3+z)6+18z?

Alex787 [66]

18 + 6z + 18z
24z + 18

4 0

3 years ago