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3 years ago

6

A packing company makes boxes with each measuring 3 feet what is the volume of the boxes boxes are put in a larger rectangle or

shipping container completely filled with no gas or overlaps what is the volume of a shipping container

1 answer:

dimulka [17.4K]3 years ago

4 0

V=L*W*H
V= 3*3*3
V= 9*3
V= 27

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The area of a right triangle is 60 ft2. The base of the triangle is 8 ft. What is the length of the hypotenuse? Show your work.

Snowcat [4.5K]

I think the hypotenuse is 17

4 0

3 years ago

Round to the nearest hundred 5,503,569

Whitepunk [10]

If the next digit, which in this case is in the tens place, is a 5 or above then we know to round up.

Your answer is 5,503,600.

Hope this helps!

5 0

3 years ago

I need help in this 10q quiz

slava [35]

Answer:

option A

Step-by-step explanation:

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3 years ago

Read 2 more answers

Find the surface area of the composite figure.

taurus [48]

Answer:

surface area is 39

Step-by-step explanation:

add the areas of each geometric figure making up the composite 3D figure.

first 3D figure

2+2+6+6

=16---eq 1

from third 3D figure

4+4+10+5

= 23

from 1 and 2

16+23

= 39

may be!! I'm not sure bout this answer

3 0

3 years ago

Pls answer the question attached

barxatty [35]

$\huge\mathfrak{\underline{answer:}}$

$\large\bf{\angle ACD = 105°}$

__________________________________________

$\large\bf{\underline{Here:}}$

BCD is an isosceles right triangle , right angled at D
ABC is an equilateral triangle

$\large\bf{\underline{To\:find:}}$

∠ ACD

$\large\bf{In\: triangle\:ABC:}$

❒ Sum of all angles is 180° , since it is an equilateral triangle all the three angles would be same

$\large\bf{\underline{So}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle ACB = \frac{180}{3}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle ACB = 60°}$

$\large\bf{In\: triangle\:BDC}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle D = 90°}$

$\bf{⟼\angle DBC =\angle DCB }$ (Isosceles triangle)

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle DBC + \angle BDC + \angle DCB = 180° }$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼2\angle DCB + 90° = 180°}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼2\angle DCB = 180 -90}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle DCB = \frac{90}{2}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle DCB = 45°}$

$\large\bf{\underline{Therefore,}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟹\angle ACD = \angle ACB + \angle DCB}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟹\angle ACD = 60+45}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟹\angle ACD = 105°}$

6 0

3 years ago