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Natalka [10]
3 years ago
10

Given a is parallel to b, angle 1 = 56°, and angle 2 = 42°, find the measure to the other angles. NOTE: Angles may not be drawn

to scale.

Mathematics
1 answer:
trasher [3.6K]3 years ago
7 0
This is going to be a pretty large answer because there are 12 angles to figure out, but here it is:
Angle 6: 56 + 42 = 98
180 - 98 = 82° = angle 6
Angle 5: 82 + 56 = 138
180 - 138 = 42° = angle 5
Angle 4: 82 + 42 = 124
180 - 124 = 56° = angle 4
Angle 3: 42 + 56 = 98
180 - 98 = 82° = angle 3
-
Angle 7: 180 - 82 (angle 6) = 98° = angle 7 [C angle]
Angle 8: 180 - 98 = 82° = angle 8
Angle 9: 180 - 82 = 98° = angle 9
Angle 10: 180 - 98 = 82° = angle 10
-
Angle 11: 82 (angle 8) + 82 (angle 6) = 164
180 - 164 = 16° = angle 11
[Angles in triangle add up to 180]
Angle 12: 180 - 16 = 164° = angle 12
Angle 13: 180 - 164 = 16° = angle 13
Angle 14: 180 - 16 = 164° = angle 14
-
Let me know if I missed one! Hope I helped!
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Answer:

1,3,1

Step-by-step explanation:

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3 years ago
Four times a number added to 3 times a larger number is 31. Seven subtracted from the larger number is equal to twice the smalle
aev [14]
Let the smaller number be x.
Let the bigger number be y.

\begin{cases} &4x + 3y = 31 \tex{ ----- (1) } \\ &y- 7 = 2x \tex{ ----- (2) } \end{cases}

Rearrange equation (2):
\begin{cases} &4x + 3y = 31 \tex{ ----- (1) } \\ &y = 2x + 7 \tex{ ----- (2) } \end{cases}



Sub (2) into (1):

4x + 3(2x + 7 ) = 31
4x + 6x + 21 = 31
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Sub x = 1 into (2):

y = 2x + 7
y = 2(1) + 7
y= 9

Answer: The two numbers are 1 and 9.

5 0
3 years ago
Read 2 more answers
A basin is filled by two pipes in 12 minutes and 16 minutes respectively. Due to the obstruction of water flow after the two pip
olasank [31]

Answer:

The time duration of the two pipes restricted flow before the flow became normal is 4.5 minutes

Step-by-step explanation:

The given information are;

The time duration for the volume, V, of the basin to be filled by one of the pipe, A, = 12 minutes

The time duration for the volume, V, of the basin to be filled by the other pipe, B, = 16 minutes

Therefore, the flow rate of pipe A = V/12

The flow rate of pipe B = V/16

Due to the restriction, we have;

The proportion of its carrying capacity the first pipe, A, carries = 7/8 of the carrying capacity

The proportion of its carrying capacity the second pipe, B, carries = 5/6 of the carrying capacity

Whereby the tank is filled 3 minutes after the restriction is removed, we have;

\dfrac{7}{8} \times \dfrac{V}{12} \times t + \dfrac{5}{6} \times \dfrac{V}{16} \times t +  \dfrac{V}{12} \times 3 + \dfrac{V}{16} \times 3 = V

Simplifying gives;

\dfrac{(2\cdot t +7) \cdot V}{16}  = V

2·t + 7 = 16

t = (16 - 7)/2 = 4.5 minutes

Therefore, it took 4.5 seconds of the restricted flow before the the flow of water in the two pipes became normal

7 0
4 years ago
Read 2 more answers
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Hi

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3 years ago
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