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Mathematics

1 answer:

olganol [36]2 years ago

4 0

Answer:

Yes, SAS theorem. Angle ACB and Angle ECD are congruent because of the vertical angles theorem.

Step-by-step explanation:

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The internal height of a rectangular box is 10.5cm.The thickness of the bottom is 3/5cm and the thickness of the top is 1cm.What

omeli [17]

The addition is one of the four fundamental mathematical operations, the others being subtraction, multiplication, and division. The external height of the box is 12.1 cm.

<h3>What is Addition?</h3>

The addition is one of the four fundamental mathematical operations, the others being subtraction, multiplication, and division. When two whole numbers are added together, the total quantity or sum of those values is obtained.

Given that the internal height of a rectangular box is 10.5cm. The thickness of the bottom is 3/5cm and the thickness of the top is 1cm. Therefore, the external height of the rectangular box is,

Height of rectangular box

= Thickness of bottom + Internal height of the box + Thickness of the top

= 3/5 cm + 10.5 cm + 1 cm

= 0.6cm + 10.5cm + 1cm

= 12.1 cm

Hence, the external height of the box is 12.1 cm.

Learn more about Addition here:

brainly.com/question/14092461

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7 0

1 year ago

13.

liubo4ka [24]

Answer:

6x +4 where x is amount for standard call

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

6) the circular base of a hemisphere of radius 2 rests on the base of a square pyramid of height 6. the hemisphere is tangent to

Rasek [7]

The length of the square base is thus $2 x 3\sqrt{2}/2 = 3\sqrt{2}$ = A

<h3>What is hemisphere?</h3>

Consequently, a hemisphere is a 3D geometric object that is made up of half of a sphere, with one side being flat and the other being a bowl-like shape. It is created by precisely cutting a spherical along its diameter, leaving behind two identical hemispheres.

EXPLANATION; Let ABCDE be the pyramid with ABCD as the square base. Let O and M be the center of square ABCD and the midpoint of side AB respectively. Lastly, let the hemisphere be tangent to the triangular face ABE at P.

Notice that triangle EOM has a right angle at O. Since the hemisphere is tangent to the triangular face ABE at P, angle EPO is also 90 degree. Hence, triangle EOM is similar to triangle EPO.

OM/2 = 6/EP

OM = 6/EP x 2

OM = $6\sqrt{6^2 - 2^2} x 2 = 3\sqrt{2}/2}$