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Dmitriy789 [7]
3 years ago
6

Find the volume of the Canada post mailbox shown below.

Mathematics
1 answer:
harina [27]3 years ago
6 0

Answer:

1,152,112.5 cm3

Step-by-step explanation:

To find the volume of this mailbox we can separate in two parts:

A rectangular prism, with dimensions 180 cm x 70 cm x 85.5 cm and

A triangular prism, with dimensions 25 cm x 70 cm x 85.5 cm.

Now we need to find the volume of both parts:

V_rectangular = 180 * 70 * 85.5 = 1,077,300 cm3

V_triangular = (1/2) * 25 * 70 * 85.5 = 74,812.5 cm3

Finally, we just need to sum the volumes of the parts:

V_mailbox = V_rectangular + V_triangular = 1077300 + 74812.5 = 1,152,112.5 cm3

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V=4/3 3.14 18*18*18 what is the volume of the sphere
podryga [215]

Answer:

24416.64

Step-by-step explanation:

Based on the standard formula,4/3×π×r^3

The π has been given already and r=18

4/3×3.14×18^3

V=24416.64

8 0
3 years ago
Choose the letter of the correct answer (Complete the congruence statement)
rewona [7]

Answer:

D

Step-by-step explanation:

when flipping and turning the second triangle, so that it is oriented as the first triangle (we see DE corresponds to AB - consider at what corner point the right angle is : at A and at D; DF corresponds to AC, EF corresponds to BC), we see immediately that BCA is the same sequence of corners as the corresponding EFD.

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Question 2 (1 point)<br> How much does the prefix milli multiply the value of a base unit?
I am Lyosha [343]

Answer:1000

Step-by-step explanation:

                                     

7 0
3 years ago
In the kite, the orange diagonal is 10 units and the purple diagonal is 14 units. What is the area of the kite?
Vlada [557]

Answer: 70 square units

Step-by-step explanation:

Area of kite = (one diagonal * another diagonal) / 2

10 * 14 = 140

140 / 2 = 70

3 0
3 years ago
What is the distance from point (– 1, 3) to the line 3x – 4y = 10?​
natta225 [31]

Answer:

5 units

Step-by-step explanation:

The distance from a point (m, n ) to the line Ax + By + C = 0 is given by

d = \frac{|Am+Bn+C|}{\sqrt{A^2+B^2} }

For the point (- 1, 3 ) , with m = - 1 and n = 3

3x - 4y = 10 ( subtract 10 from both sides )

3x - 4y - 10 = 0

with A = 3 , B = - 4 , C = - 10

d = \frac{|3(-1)+(-4)3-10|}{\sqrt{3^2+(-4)^2} }

   = \frac{|-3-12-10|}{9+16}

   = \frac{|-25|}{\sqrt{25} }

    = \frac{25}{5}

     = 5

5 0
2 years ago
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