2.1.4 Practice: Adding Rational Numbers

A fruit stand sells apples, peaches and tomatoes. Today, they sold 4 apples for every 5 peaches. They sold 2 peaches for every 3

Brrunno [24]

Answer:

He sold 32 apples, 40 peaches and 60 apples.

Step-by-step explanation:

We are given that

4 apples are sold every 5 peaches and

3 tomatoes are sold every 2 peaches.

i.e. the ratio is apples: peaches = 4 : 5 ...... (1)

and the ratio peaches: tomatoes = 2 : 3 ...... (2)

Peaches are the common link here in both the ratio terms.

Let us make the peaches value in both the <em>ratio as 10 </em>to make it equal in both the ratio terms.

Multiply (1) by 2 and (2) by 5:

The ratio becomes;

apples:peaches = 8:10

peaches: tomatoes = 10:15

In other words, the ratio becomes:

apples:peaches:tomatoes = 8:10:15

Let the number of apples = $8x$

Let the number of peaches = $10x\\$

Let the number of tomatoes = $15x$

We are given that total fruits are = 132

$\Rightarrow 8x+10x+15x = 132\\\Rightarrow 33x = 132\\\Rightarrow x =4\\$

So, number of apples sold = $8 \times 4 = 32$

Number of peaches sold = $10 \times 4 = 40$

Number of tomatoes sold = $15 \times 4 = 60$

7 0

3 years ago

2/3+1/6???????????????????

kupik [55]

Answer:

5/6

Step-by-step explanation:

3 0

4 years ago

Read 2 more answers

Https://forms.gle/jKAg2FoMP8HFPiLR7<br> Do this if you want too.

Katen [24]

Answer:

You need permission to do that.

Step-by-step explanation:

8 0

3 years ago

Pls help me so I can pass

spin [16.1K]

Answer:

Sorrry i dont know

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Use the long division method to find the result when x^3+9x² +21x +9 is divided<br> by x+3

Serhud [2]

Answer:

x^3 + 9 x^2 + 21 x + 9 = (x^2 + 6 x + 3)×(x + 3) + 0

Step-by-step explanation:

Set up the polynomial long division problem with a division bracket, putting the numerator inside and the denominator on the left:

x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9

To eliminate the leading term of the numerator, x^3, multiply x + 3 by x^2 to get x^3 + 3 x^2. Write x^2 on top of the division bracket and subtract x^3 + 3 x^2 from x^3 + 9 x^2 + 21 x + 9 to get 6 x^2 + 21 x + 9:

| | | x^2 | | | |

x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9

| -(x^3 | + | 3 x^2) | | | |

| | | 6 x^2 | + | 21 x | + | 9

To eliminate the leading term of the remainder of the previous step, 6 x^2, multiply x + 3 by 6 x to get 6 x^2 + 18 x. Write 6 x on top of the division bracket and subtract 6 x^2 + 18 x from 6 x^2 + 21 x + 9 to get 3 x + 9:

| | | x^2 | + | 6 x | |

x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9

| -(x^3 | + | 3 x^2) | | | |

| | | 6 x^2 | + | 21 x | + | 9

| | | -(6 x^2 | + | 18 x) | |

| | | | | 3 x | + | 9

To eliminate the leading term of the remainder of the previous step, 3 x, multiply x + 3 by 3 to get 3 x + 9. Write 3 on top of the division bracket and subtract 3 x + 9 from 3 x + 9 to get 0:

| | | x^2 | + | 6 x | + | 3

x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9

| -(x^3 | + | 3 x^2) | | | |

| | | 6 x^2 | + | 21 x | + | 9

| | | -(6 x^2 | + | 18 x) | |

| | | | | 3 x | + | 9

| | | | | -(3 x | + | 9)

| | | | | | | 0

The quotient of (x^3 + 9 x^2 + 21 x + 9)/(x + 3) is the sum of the terms on top of the division bracket. Since the final subtraction step resulted in zero, x + 3 exactly divides x^3 + 9 x^2 + 21 x + 9 and there is no remainder.

| | | x^2 | + | 6 x | + | 3 | (quotient)

x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9 |

| -(x^3 | + | 3 x^2) | | | | |

| | | 6 x^2 | + | 21 x | + | 9 |

| | | -(6 x^2 | + | 18 x) | | |

| | | | | 3 x | + | 9 |

| | | | | -(3 x | + | 9) |

| | | | | | | 0 | (remainder) invisible comma

(x^3 + 9 x^2 + 21 x + 9)/(x + 3) = (x^2 + 6 x + 3) + 0

Write the result in quotient and remainder form:

Answer: Set up the polynomial long division problem with a division bracket, putting the numerator inside and the denominator on the left:

x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9

To eliminate the leading term of the numerator, x^3, multiply x + 3 by x^2 to get x^3 + 3 x^2. Write x^2 on top of the division bracket and subtract x^3 + 3 x^2 from x^3 + 9 x^2 + 21 x + 9 to get 6 x^2 + 21 x + 9:

| | | x^2 | | | |

x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9

| -(x^3 | + | 3 x^2) | | | |

| | | 6 x^2 | + | 21 x | + | 9

To eliminate the leading term of the remainder of the previous step, 6 x^2, multiply x + 3 by 6 x to get 6 x^2 + 18 x. Write 6 x on top of the division bracket and subtract 6 x^2 + 18 x from 6 x^2 + 21 x + 9 to get 3 x + 9:

| | | x^2 | + | 6 x | |

x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9

| -(x^3 | + | 3 x^2) | | | |

| | | 6 x^2 | + | 21 x | + | 9

| | | -(6 x^2 | + | 18 x) | |

| | | | | 3 x | + | 9

To eliminate the leading term of the remainder of the previous step, 3 x, multiply x + 3 by 3 to get 3 x + 9. Write 3 on top of the division bracket and subtract 3 x + 9 from 3 x + 9 to get 0:

| | | x^2 | + | 6 x | + | 3

x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9

| -(x^3 | + | 3 x^2) | | | |

| | | 6 x^2 | + | 21 x | + | 9

| | | -(6 x^2 | + | 18 x) | |

| | | | | 3 x | + | 9

| | | | | -(3 x | + | 9)

| | | | | | | 0

The quotient of (x^3 + 9 x^2 + 21 x + 9)/(x + 3) is the sum of the terms on top of the division bracket. Since the final subtraction step resulted in zero, x + 3 exactly divides x^3 + 9 x^2 + 21 x + 9 and there is no remainder.

| | | x^2 | + | 6 x | + | 3 | (quotient)

x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9 |

| -(x^3 | + | 3 x^2) | | | | |

| | | 6 x^2 | + | 21 x | + | 9 |

| | | -(6 x^2 | + | 18 x) | | |

| | | | | 3 x | + | 9 |

| | | | | -(3 x | + | 9) |

| | | | | | | 0 | (remainder) invisible comma

(x^3 + 9 x^2 + 21 x + 9)/(x + 3) = (x^2 + 6 x + 3) + 0

Write the result in quotient and remainder form:

Answer: x^3 + 9 x^2 + 21 x + 9 = (x^2 + 6 x + 3)×(x + 3) + 0

5 0

2 years ago