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madreJ [45]
3 years ago
9

max has 6 cartons. in each carton,he places 6 bags. in each of the bags, he places 6 cookies. how many cookies are contained in

the cartons? use an exponent to write the answer

Mathematics
1 answer:
Elenna [48]3 years ago
6 0
6^3 (6 to the third power) because it is 6 x 6 x 6
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Please help i got a f on it and i am redoing it i will give 28 points for the right answer
Savatey [412]

Answer:

11/16

Step-by-step explanation:

12 divided by (2+2/3)^2 =1.6875

1.6875 in a decimal form is 11/16

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2 years ago
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Find the midpoint between the two points (-4,4) and (-2,2)
Ymorist [56]

Answer: (-3,3) is the answer

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5/7^3=5(7^x)<br> Find X.<br> Thanks.
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Hello,

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8 0
3 years ago
A metal hollow bar whose cross section and dimension are shown below weighs 8x10^3 kg/m^3 and measure 2m in length ..determine t
devlian [24]
1) We calculate the volume of a metal bar (without the hole).

volume=area of hexagon x length
area of hexagon=(3√3 Side²)/2=(3√3(60 cm)²) / 2=9353.07 cm²
9353.07 cm²=9353.07 cm²(1 m² / 10000 cm²)=0.935 m²

Volume=(0.935 m²)(2 m)=1.871 m³

2) we calculate the volume of the parallelepiped

Volume of a parallelepiped= area of the section  x length
area of the section=side²=(40 cm)²=1600 cm²
1600 cm²=(1600 cm²)(1 m² / 10000 cm²=0.16 m²
Volume of a parallelepiped=(0.16 m²)(2 m)=0.32 m³

3) we calculate the volume of a metal hollow bar:
volume of a metal hollow bar=volume of a metal bar -  volume of a parallelepiped

Volume of a metal hollow bar=1.871 m³ - 0.32 m³=1.551 m³

4) we calculate the mass of the metal bar

density=mass/ volume  ⇒ mass=density *volume

Data:
density=8.10³ kg/m³
volume=1.551 m³

mass=(8x10³ Kg/m³ )12. * (1.551 m³)=12.408x10³ Kg

answer: The mas of the metal bar is 12.408x10³ kg  or   12408 kg  


4 0
3 years ago
Mr. Kohl has a breaker containing n milliliters of solution to distribute to the students in his chemistry class. If he gives ea
77julia77 [94]

Answer:

There were 26 students in his class and the teacher had 83 ml of the solution.

Step-by-step explanation:

Mr. Kohl has a "x" amount of solution, if he divides it by the number of students "n" he'll give each student 3 milliliters  and have a left over of 5 milliliters. If the amount of solution Mr. Kohl had was "x + 21" then he'd be able to give each student 4 milliliters of the solution. From these informations we have:

x = 3*n + 5

(x + 21)/n = 4

x + 21 = 4*n

x = 4*n - 21

Now that we have two equations and two variables we can solve the system of equations, as seen bellow:

3*n + 5 = 4*n - 21

3*n - 4*n = -21 - 5

-n = -26

n = 26

x = 4*26 - 21 = 83 ml

There were 26 students in his class and the teacher had 83 ml of the solution.

3 0
3 years ago
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