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sveta [45]
3 years ago
8

What dimensions of the box are required to fit the candle? 5 by 6 by 18 7 by 6 by 18 5 by 3 by 18 7 by 3 by 18

Mathematics
1 answer:
LiRa [457]3 years ago
4 0
The minimum height of the box must be 18 units, which is present in all the choices.
Another minimum dimension that we already know is 7 units, since it is an oblique box, and the projection of bottom-left to top-right is shown as 7 units.
That leaves us with the second (7x6) or fourth choice (7x3).

We have not been given the other dimension of the triangle, but from common sense, it looks longer than the front (5 units) leg, therefore 6 units is not an unreasonable dimension.

With 99.5% certainty, I would choose the second choice (7x6x18) because the remaining dimension does not look like 3 units long.
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The answer is A, 14 inches.

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2 complementary angles are in the ratio 7:8.find the angles
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Answer:

1) ratio

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3 0
3 years ago
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pentagon [3]

Answer:

B = 34.2°

C = 58.2° or 121.8°

c= 10.6

Step-by-step explanation:

Step 1

Finding c

We calculate c using Pythagoras Theorem

c²= a² + b²

c = √a² + b²

a= 8, b = 7

c = √8² + 7²

c = √64 + 49

c = √(113)

c = 10.630145813

Approximately c = 10.6

Step 2

Find B

We solve this using Sine rule

a/sin A = b/sin B

A = 40°

a = 8

b = 7

Hence,

8/sin 40° = 7/sin B

8 × sin B = sin 40° × 7

sin B = sin 40° × 7/8

B = arc sin (sin 40° × 7/8)

B ≈34.22465°

Approximately = 34.2°

Step 3

We find C

Find B

We solve this using Sine rule

b/sin B = c/sin C

B = 34.2°

b = 7

c = 10.6

C = ?

Hence,

7/sin 34.2° = 10.6/sin C

7 × sin C = sin 34.2 × 10.6

sin C = sin 34.2° × 10.6/7

C = arc sin (sin 34.2° × 10.6/7)

C = arcsin(0.85)

C= 58.211669383

Approximately C = 58.2°

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Answer:

0.00539218411

Step-by-step explanation:

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3 years ago
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