Does anyone know this?

A bag of marbles can be divided evenly among 2,3 or 4 friends. A) How many marbles might be in the bag? B) What is the least num

agasfer [191]

Answer:

a)12 b)12 c)36

Step-by-step explanation:

a)find least common multiple. the number of marbles is that number * n (where n is an integer)

b) the least common multiple. it is 4*3=...

c)12*3 =?

then just divide by 2,3,4 to get answer

6 0

3 years ago

Lee las situaciones y realiza lo siguiente con cada una:

Julli [10]

Answer:

Part 1) see the explanation

Part 2) see the explanation

Part 3) see the explanation

Part 4) see the explanation

Step-by-step explanation:

The question in English is

Read the situations and do the following with each one:

Write down the magnitudes involved

Write which magnitude is the independent variable and which is the dependent variable

It represents the function that describes the situation

SITUATIONS:

1) A machine prints 840 pages every 30 minutes.

2) An elevator takes 6 seconds to go up two floors.

3) A company rents a car at S/ 480 for 12 days.

4) 10 kilograms of papaya cost S/ 35

Part 1) we have

A machine prints 840 pages every 30 minutes

Let

x ----> the time in minutes (represent the variable independent or input value)

y ---> the number of pages that the machine print (represent the dependent variable or output value)

Remember that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $k=\frac{y}{x}$ or $y=kx$

In this problem

we have a a proportional variation

so

The value of the constant of proportionality is equal to

$k=\frac{y}{x}$

we have

$y=840\ pages\\x=30\ minutes$

substitute

$k=\frac{840}{30}=28\ pages/minute$

The linear equation is

$y=28x$

Part 2) we have

An elevator takes 6 seconds to go up two floors.

Let

x ----> the time in seconds (represent the variable independent or input value)

y ---> the number of floors (represent the dependent variable or output value)

Remember that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $k=\frac{y}{x}$ or $y=kx$

In this problem

we have a a proportional variation

so

The value of the constant of proportionality is equal to

$k=\frac{y}{x}$

we have

$y=2\ floors\\x=6\ seconds$

substitute

$k=\frac{2}{6}=\frac{1}{3}\ floors/second$

The linear equation is

$y=\frac{1}{3}x$

Part 3) we have

A company rents a car at S/ 480 for 12 days.

Let

x ----> the number of days (represent the variable independent or input value)

y ---> the cost of rent a car (represent the dependent variable or output value)

Remember that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $k=\frac{y}{x}$ or $y=kx$

In this problem

we have a a proportional variation

so

The value of the constant of proportionality is equal to

$k=\frac{y}{x}$

we have

$y=\$480\\x=12\ days$

substitute

$k=\frac{480}{12}=\$40\ per\ day$

The linear equation is

$y=40x$

Part 4) we have

10 kilograms of papaya cost S/ 35

Let

x ----> the kilograms of papaya (represent the variable independent or input value)

y ---> the cost (represent the dependent variable or output value)

Remember that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $k=\frac{y}{x}$ or $y=kx$

In this problem

we have a a proportional variation

so

The value of the constant of proportionality is equal to

$k=\frac{y}{x}$

we have

$y=\$35\\x=10\ kg$

substitute

$k=\frac{35}{10}=\$3.5\ per\ kg$

The linear equation is

$y=3.5x$

6 0

3 years ago

If you are smart put this emojiif u not smart put this emoji ok thanks

PtichkaEL [24]

Yesssss that is right

3 0

4 years ago

Read 2 more answers

How would u find 20% of the regular price? what would you do with this amount to find the sale price

hjlf

Answer:

you would multiply the price by .20

Step-by-step explanation:

example:

20% of $160

sale, 20% off

160 x .20 = 32

$32 is 20% of $160, $128 is the sale price

3 0

3 years ago

In a single growing season at the smith family orchard, the average yield per apple tree is 150 apples when the number of trees

Mariana [72]

Answer:

100 trees should be planted per acre to maximize the total yield (15000 apples per acre)

Step-by-step explanation:

Functions

Expressing some quantities in terms of other(s) comes in handy to understand the behavior of the situations often modeled by those functions.

Our problem tells us about the smith family orchard, where the average yield per apple tree is 150 apples when the number of trees per acre is 100 (or less). But the average yield per tree decreases when more than 100 trees are planted. The rate of decrease is 1 for each additional tree is planted

Let's call x the number of trees planted per acre in a single growing season. We have two conditions when modeling: when x is less than 100, the average yield per apple tree (Ay) is (fixed) 150.

$Ay(x)=150\ ,\ x\leqslant 100$

When x is more than 100:

$Ay(x)=150-x\ ,\ 100< x\leqslant 150$

The upper limit is 150 because it would mean the Average yield is zero and no apples are to be harvested

Part 1

We are asked what happens if the number of trees per acre was doubled, but we don't know what the original number was. Let's make it x=60

If x=60, then we must use the first function, and

Ay(60)=150

The yield per acre will be 150*60=9000 apples

If we doubled x to 120, we would have to use the second function

Ay(120)=150-x=30

The yield per acre will be 30*120=3600 apples

Note that doubling the number of trees would make the production decrease

If we started with x=30, then

Ay(30)=150, and we would have 150*30=4500 apples per acre

We already know that for x=60 (the double of 30), our production will be 9000 apples per acre

This shows us that in some cases, increasing the number of trees is beneficial and in other cases, it is not

Part 2

The total yield is computed as

Y(x)=x.Ay(x)

This function will be different depending on the value of x

$Y(x)=150x\ ,\ x\leqslant 100$

$Y(x)=150x-x^2\ ,\ 100< x\leqslant 150$

To find the maximum total yield we take the first derivative

$Y'(x)=150\ ,\ x\leqslant 100$

$Y'(x)=150-2x\ ,\ 100< x\leqslant 150$

The first expression has no variable, so the total yield must be evaluated in the endpoints

Y(0)=0

Y(100)=15000

Now, the second expression contains variable, it will be set to 0 to find the critical point

150-2x=0

x=75

x is outside of the boundaries, we cannot find the maximum in this interval

Answer: 100 trees should be planted per acre to maximize the total yield (15000 apples per acre)

7 0

4 years ago