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2 years ago

A path is 3 feet wide and is paved Weight square brakes that are 3 inches wide. How many bricks wide is the path

1 answer:

4 0

The area of the path is can be solved using the equation below:
Area path = wide²
First, let us convert 3 feet to inches such as shown below:
3 feet * (12 inches / 1 foot) = 36 inches
Area of the path = 36²
Area of the path = 1296 inches²

Solving for the area of the square bricks
Area of the bricks = (3 inches)²
Area of the brick = 9 inches²

Solving for the number of bricks, we have it:
Total number of bricks = Area of the path/area of the bricks
Total number of bricks = 1296 / 9
Total number of bricks = 144 pieces

The answer is 144 total number of bricks.

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Find the difference 0.2 - 0.05 =

Ugo [173]

Answer:

0.2-0.05=0.15

Step-by-step explanation:

0.2=0.20

0.20-0.05=0.15

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2 years ago

The sum of all integers between 3 and 11

Lena [83]

Answer:

Step-by-step explanation:

You would add 4+5+6+7+8+9+10 to get the sum of 49.

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2 years ago

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Julia has 14 pounds of nuts. There are 16 ounces in 1 pound. How many ounces of nuts does she have?

AlladinOne [14]

She has 224 ounces of nuts

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2 years ago

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Help please.......Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.

maxonik [38]

Answer:

The cost of 1 dozen doughnut is $6 and the cost of a dozen croissant is $10.

Step-by-step explanation:

We know that:

D = Cost of a dozen doughnut
C = Cost of a dozen croissant
10D + 7C = 130
6D + 4C = 76

Work:

10D + 7C = 130

6D + 4C = 76

4(10D + 7C = 130)

7(6D + 4C = 76)

40D + 28C = 520

42D + 28C = 532

40D - 42D = 520 - 532
=> -2D = -12
=> 2D = 12
=> D = 6

Now, let's substitute the value of D into the first equation.

10D + 7C = 130
=> 10(6) + 7C = 130
=> 60 + 7C = 130
=> 7C = 70
=> C = 10

Hence, the cost of 1 dozen doughnut is $6 and the cost of a dozen croissant is $10.

Hoped this helped.

$BrainiacUser1357$

4 0

1 year ago

Will give brainliest to best answer.

Tomtit [17]

Problem One
Remark
If she doesn't mind having I kg left over, the minimum number would be 3 five kg boxes. If on the other hand, she must have exactly 14 kg then the minimum number is 6.

She needs 2 five kg boxes and 4 one kg boxes. <<<< Answer

Problem Two
There is a method of solving this that is called dimensional analysis. It is what should be used here. I'll do it at the end of the problem. In the meantime, you have to do it a slightly longer way.

1 portion = 100 grams.
x portions = 1kg which is 1000 grams.
x portions = 1000 grams.

Set up a proportion to find the number of servings in 1 kg
1 portion/x = 100 grams/1000 grams Cross multiply
1 * 1000 = 100 * x Divide by 100
1000/100 = x
x = 10 servings in 1 kg.

So each kg produces 10 portions
1 kg / 10 portions = 20 kg / x portions Cross multiply
x * 1 = 10 * 20
x = 200 portions <<<<< Answer

Dimensional Analysis
[1batch]*[1 portion/100g][1000g/kg][20kg/batch] the units cancel
1000 * 20 / 100 only the portions are left over.
200 portions is the answer.

Problem Three
1 kg = 1000 grams
x kg = 5000 grams Cross multiply

1*5000 = 1000 x
x = 5 kg

1 parcel weighs 5 kg
x = 15 kg

15 kg = 5 x
x = 15/5
x = 3

So he can carry 3 parcels per trip.

Since there are 5 such parcels, he will have to make 2 trips. The second one will not be a full load.
First Trip = 3 parcels
Second Trip = 2 parcels. <<<<Answer

5 0

2 years ago