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ira [324]
3 years ago
15

In this figure, mBDA=___ and mBCA=___.

Mathematics
2 answers:
Serhud [2]3 years ago
5 0
∠BDA is inscribed angle
inscribed angle = 1/2 × central angle
the central angle is small ∠BOA, use ∠BOA that face the same direction as ∠BDA

Find the central angle ∠BOA
central angle = 360° - large ∠BOA
central angle = 360° - 250°
central angle = 110°

Find ∠BDA
inscribed angle = 1/2 × central angle
∠BDA = 1/2 × small ∠BOA
∠BDA = 1/2 × 110°
∠BDA = 55°


To find ∠BCA, use tetragon BOAC to solve the problem
The sum of interior angles in tetragon is 360°. Because CA and CB is tangent of the circle, the angles which is formed by the tangent is 90°. So ∠CAO and ∠CBO are both equal to 90°
∠BCA + ∠CAO + ∠CBO + small ∠BOA = 360°
∠BCA + 90° + 90° + 110° = 360°
∠BCA + 290° = 360°
∠BCA = 360° - 290°
∠BCA = 70°

THE ANSWER
m∠BDA = 55°
m∠BCA = 70°
san4es73 [151]3 years ago
4 0
MBDA= ADB and mBCA= ACB
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that is b.
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Name each polynomial<br><br> 4m^2-5m+6<br><br> 18s-7<br><br> 6p^3
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2 years ago
A padlock has the digits 0 through 9 arranged in a circle on its face. A combination for 0 1 2 3 4 6 5 7 8 9 this padlock is fou
stepan [7]

Answer:

The total numbers of possible combinations are 3430.

Step-by-step explanation:

Consider the provided information.

A combination for 0 1 2 3 4 6 5 7 8 9 this padlock is four digits long. Because of the internal mechanics of the lock, no pair of consecutive numbers in the combination can be the same or one place apart on the face.

Here, for the first digit we have 10 choices.

For the second digit we have 7 choices, as the digit can't be the same nor adjacent to the first digit.

For the third digit we have 7 choices, as the digit can't be the same nor adjacent to the second digit.

For the fourth digit we have 7 choices, as the digit can't be the same nor adjacent to the third digit.

So the number of choices are:

10\times 7\times 7\times 7=10\times 7^3\\10\times 343=3430

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8 0
2 years ago
A manager in a call center needs to determine how many agents to schedule next week.
lukranit [14]

Answer:

Number of calls expected in next week by manager = 7940

Average Number of calls that call center agent will attend in an hour =7 calls

It is also given that, Call center remain open for 10 hours 5 days a week.

Also, it is given that, full time agents work 40 hours a week but are only on call for 35 hours per week  ,Part time agents work 20 hours a week but are only on calls 17 hours per week .

⇒Number of hours worked by full time agents × Number of calls attended in an hour × Number of full time agents + Number of hours worked by Part time agents × Number of calls attended in an hour × Number of Part time agents ≤ 7940

⇒35 × 7×Number of full time agents +17 × 7 ×Number of Part time agents ≤ 7940

Option A

⇒35×15×7+17×7×15

= 3675+1785

= 5460

Option B

⇒35 ×7×20+17×7×7

=4900 +833

= 5733

Option C

⇒35×20×7 +17×20×7

=4900+2380

=7280

Option D

⇒25 × 35×7+17×7×5

=6125 +595

=6720

Option E

⇒28×35×7+17×7×10

=6860+1190

=8050

Option E, ⇒ 28 full time agents and 10 part time agents , is best to meet the scheduling needs is most appropriate, that is nearer to 7940 calls.

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2 years ago
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