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Mashutka [201]
3 years ago
14

The size of the angles of a quadrilateral are in the ratio

Mathematics
1 answer:
Irina-Kira [14]3 years ago
8 0

Answer:

40°, 80°, 120°, 120°

Step-by-step explanation:

sum the parts of the ratio, 1 + 2 + 3 + 3 = 9 parts

The sum of the interior angles of a quadrilateral is 360°

Divide this by 9 to find the value of one part of the ratio.

360° ÷ 9 = 40° ← value of 1 part of the ratio

Then

2 parts = 2 × 40° = 80°

3 parts = 3 × 40° = 120°

The 4 angles are 40°, 80°, 120°, 120°

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Jesse is looking for the volume of the cone below, and she is using the formula V=(1/3)Bh. Which of the following is the value o
Sidana [21]

Answer:

B =  \pi *r^2

(Base of the cone)

Step-by-step explanation:

The volume of the cone is always 1/3 of the volume of a cylinder with the same radius and height.

Volume of the cylinder

V_cylin = ( \pi *r^2 )* h

Where

r is the radius

h is the height

This means the volume of the cone is equal to

V_cone = (1/3)* ( \pi *r^2 )* h

By looking to the equation of the problem

V=(1/3)Bh

We can easily deduce that

B =  \pi *r^2

(Base of the cone)

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What is the place value of 8 in 281480100
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The number 8 is in Ten thousand space
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What is 8,690,000 rounded to the nearest tenth.?
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Look at this set of ordered pairs:
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3 0
2 years ago
State the number of possible triangles that can be formed using the given measurements.
romanna [79]

Answer:  39) 1              40) 2

                41) 1              42) 0

<u>Step-by-step explanation:</u>

39)     ∠A = ?        ∠B = ?       ∠C = 129°

            a = ?          b = 15         c = 45

Use Law of Sines to find ∠B:

\dfrac{\sin B}{b}=\dfrac{\sin C}{c} \rightarrow\quad \dfrac{\sin B}{15}=\dfrac{\sin 129}{45}\rightarrow \quad \angle B=15^o\quad or \quad \angle B=165^o

If ∠B = 15°, then ∠A = 180° - (15° + 129°) = 36°

If ∠B = 165°, then ∠A = 180° - (165° + 129°) = -114°

Since ∠A cannot be negative then ∠B ≠ 165°

∠A = 36°        ∠B = 15°       ∠C = 129°       is the only valid solution.

40)      ∠A = 16°        ∠B = ?       ∠C = ?

             a = 15           b = ?         c = 19

Use Law of Sines to find ∠C:

\dfrac{\sin A}{a}=\dfrac{\sin C}{c} \rightarrow\quad \dfrac{\sin 16}{15}=\dfrac{\sin C}{19}\rightarrow \quad \angle C=20^o\quad or \quad \angle C=160^o

If ∠C = 20°, then ∠B = 180° - (16° + 20°) = 144°

If ∠C = 160°, then ∠B = 180° - (16° + 160°) = 4°

Both result with ∠B as a positive number so both are valid solutions.

Solution 1:  ∠A = 16°        ∠B = 144°       ∠C = 20°    

Solution 2:  ∠A = 16°        ∠B = 4°       ∠C = 160°    

41)       ∠A = ?        ∠B = 75°       ∠C = ?

             a = 7           b = 30         c = ?

Use Law of Sines to find ∠A:

\dfrac{\sin A}{a}=\dfrac{\sin B}{b} \rightarrow\quad \dfrac{\sin A}{7}=\dfrac{\sin 75}{30}\rightarrow \quad \angle A=13^o\quad or \quad \angle A=167^o

If ∠A = 13°, then ∠C = 180° - (13° + 75°) = 92°

If ∠A = 167°, then ∠C = 180° - (167° + 75°) = -62°

Since ∠C cannot be negative then ∠A ≠ 167°

∠A = 13°        ∠B = 75°       ∠C = 92°       is the only valid solution.

42)      ∠A = ?         ∠B = 119°       ∠C = ?

             a = 34         b = 34           c = ?

Use Law of Sines to find ∠A:

\dfrac{\sin A}{a}=\dfrac{\sin B}{b} \rightarrow\quad \dfrac{\sin A}{34}=\dfrac{\sin 119}{34}\rightarrow \quad \angle A=61^o\quad or \quad \angle A=119^o

If ∠A = 61°, then ∠C = 180° - (61° + 119°) = 0°

If ∠A = 119°, then ∠C = 180° - (119° + 119°) = -58°

Since ∠C cannot be zero or negative then ∠A ≠ 61° and ∠A ≠ 119°

There are no valid solutions.

6 0
3 years ago
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