There are 20 balls in a bag. Ten of the balls are red, 4 are green, and the rest are yellow.

Which ratio represents the number of yellow balls to red balls?

1:5

3:10

2:5

3:5

The expression -6m-n+3 is equivalent to our given expression.

To find the equivalent expression for -3(2m - 1) - n, distribute the -3 to 2m and -1, and then combine like terms:

[tex]\(-3(2m - 1) - n\)[/tex]

[tex]\(-6m + 3 + n\)[/tex]

So, the correct answer is option D) -6m - n + 3.

Complete question:

Which expression is equivalent to -3(2m - 1) - n?

A) 6m - n - 3

B) 6m - n + 3

C) -6m - n - 3

D) -6m - n + 3

Answer:  5

Step-by-step explanation:

Given : Tristan has 45 apples and 20 pears that he is putting into gift baskets.

Each basket will have the same number of apples and pears.

To find the greatest number of baskets Tristan can make with no fruit left over, we find the greatest common factor if 45 and 20.

For that first we write the prime factorization of 45 and 20 :-

[tex]45=3\times3\times5\\\\20=2\times2\times5[/tex]

Greatest common factor of 45 and 20 (GCF) = 5

Hence , the greatest number of baskets Tristan can make with no fruit left over =5

Answer:

Yes, we have to regroup. Why because of place value chart, we have to regroup.

Step-by-step explanation:

When we add, we have to add the numbers according to the place value chart.

Tens     ones

4            3

2             8

-----------------

 7            1

When we add ones place, we get 3 + 8 = 11, it is a two digits number, so place ones place (1) and carry over the tens (1) to the tens place.

Already, we have 4 and 2 in the tens place, another 1 tens adds to tens place.

4 + 2 + 1 = 7

Hope this will helpful.

Thank you.

To solve the equation 3x + 15y = 7 for y, subtract 3x from both sides and then divide by 15. This results in y = (7 - 3x) / 15.

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Mr. Evans can ship approximately 70 freezers in the container.

To determine how many freezers can be shipped in the container, we first need to calculate the total weight of the ovens and then subtract it from the maximum weight the container can hold.

Given:

- Maximum weight the container can hold = 42,000 pounds

- Number of ovens = 90

- Weight of each freezer = 600 pounds

First, we need to calculate the total weight of the ovens:

Total weight of ovens = Number of ovens × Weight of each oven

Total weight of ovens = 90 ovens × 0 pounds = 0 pounds

Now, we subtract the total weight of the ovens from the maximum weight the container can hold to find the remaining weight for the freezers:

Remaining weight for freezers = Maximum weight of container - Total weight of ovens

Remaining weight for freezers = 42,000 pounds - 0 pounds

Remaining weight for freezers = 42,000 pounds

Now, we need to find out how many freezers can be shipped in the container. Since each freezer weighs 600 pounds, we divide the remaining weight for freezers by the weight of each freezer:

Number of freezers = Remaining weight for freezers / Weight of each freezer

Number of freezers = 42,000 pounds / 600 pounds

Number of freezers ≈ 70

So, Mr. Evans can ship approximately 70 freezers in the container.

Answer:

The answer would be "Some number divided by 3"

Step-by-step explanation:

y = some number.

3 = 3

When an equation is shown in a fraction and you're using any terms that may be used to represent "some number," such as y, x, etc; it usually means to divide the two numbers in the fraction.

Sorry if that didn't make sense, I'm only in 7th grade and I'm very bad at explaining haha

Let's say we have the fraction [tex]\frac{1}{2}[/tex].

To turn the fraction into a decimal, you want to take the denominator and divide it into the numerator. In this case, the denominator is 2 so we will create a divison problem and it will be 1 ÷ 2 = 0.5

↑ This method will always work ↑

Let's say we have the decimal 0.3 and we want to turn that decimal into a fraction. You can see that there is 1 digit to the right of the decimal which means that the 3 is located in the tenths place. What you want to do is put the decimal over 1 so [tex]\frac{0.3}{1}[/tex] and then multiply the top and the bottom by 10. If there is two digits after the decimal point, multiply it by 100 and if there is 3, multiply it by 1,000.

[tex]\frac{0.3}{1}[/tex] × [tex]\frac{10}{10}[/tex] = [tex]\frac{3}{10}[/tex].

The answer is B. Hope this helped

Suppose 6 people can plant 12 trees in 3 hours. So, The people are needed to plant 28 trees in 5 hours and 15 minutes is 8.

The unitary method is a method for solving a problem by the first value of a single unit and then finding the value by multiplying the single value.

The time, t, in hours, that it takes x people to plant n trees varies directly

with the number of trees, and inversely with the number of people.

[tex]t \propto n\\\\t \propto \dfrac{1}{x}[/tex]

Suppose 6 people can plant 12 trees in 3 hours.

[tex]3 = 6k\\\\k = 1/2\\\\[/tex]

So, the people are needed to plant 28 trees in 5 hours and 15 minutes

= 8

Learn more about the unitary method;

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Answer:

13/20

Step-by-step explanation:

1

4

+

2

5

=  

1 × 5

4 × 5

+

2 × 4

5 × 4

=  

5

20

+

8

20

=  

5 + 8

20

=  

13

20

hope this helps

To find the width of the park, divide the area (25 / 54 square miles) by the length (5 / 9 miles), resulting in a width of

5 / 6 miles.

The question asks to find the width in miles of a rectangular city park whose area is 25 / 54 square miles and whose length is 5 / 9 miles. To find the width, we can divide the area by the length:

Width = Area / Length

Width = (25 / 54) [tex]\div[/tex] (5 / 9)

To divide fractions, we multiply by the reciprocal of the divisor:

Width = (25 / 54) X (9 / 5)

Width = (25 X 9) [tex]\div[/tex] (54 X 5)

Width = (225) [tex]\div[/tex] (270)

Then we reduce the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 45:

Width = (225 / 45) [tex]\div[/tex] (270 / 45)

Width = 5 / 6

Therefore, the width of the park is 5/6 miles.

The number of partial products in multiplication is determined by the number of digits in the multiplier.

In multiplication, the number of partial products is determined by the number of digits in the multiplier. In part a, there are three digits in the multiplier, so there are three partial products. Each partial product represents the product of a digit in the multiplier and the multiplicand. In part b, there are only two digits in the multiplier, so there are only two partial products.

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