Answer:
x = 82°
y = 59°
z = 39°
Step-by-step explanation:
∠x = 1/2(134° + 30°) = 82°
∠z = 180° - 59° - 82° = 39°
arc DF = 2(39°) = 78°
arc DB + arc DF = 134° + 78° = 212°
draw a diameter down from point B to the opposite side of the circle, call that point P:
arc BDP = 180°
(arc BDF) 212° - 180° = 32° (arc PF)
∠PBG = 1/2(32°) = 16°
∠DEB = 1/2(134°) = 67°
∠DBE = 180° - 59° - 67° = 54°
(∠DBE) 54° - (∠z) 39° + (∠PBG) 16° = (∠PBE) 31°
y = 90°- 31° = 59°
The probability that the result that is gotten will be a multiple of 3 and 2 will be 2/15.
<h3>How to calculate the probability</h3>
From the information, the spinner has 15 equal areas, numbered 1 through 15.
The multiple of 3 and a multiple of 2 among the numbers will be 6 and 15. Therefore, the probability will be two out of fifteen. This will be 2/15.
Learn more about multiples on:
brainly.com/question/1067440
Answer:
A = 52°, a = 149.2, c = 174.3
Step-by-step explanation:
Technology is useful for this. Many graphing calculators can solve triangles for you. The attachment shows a phone app that does this, too.
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The Law of Sines can give you the value of c, so you can choose the correct answer from those offered.
c = sin(C)·b/sin(B) = sin(113°)·49/sin(15°) ≈ 174.271 ≈ 174.3 . . . . . third choice
4 (i literally had to think about this cause i thought it was too easy )
Answer and Step-by-step explanation:
(a) Given that x and y is even, we want to prove that xy is also even.
For x and y to be even, x and y have to be a multiple of 2. Let x = 2k and y = 2p where k and p are real numbers. xy = 2k x 2p = 4kp = 2(2kp). The product is a multiple of 2, this means the number is also even. Hence xy is even when x and y are even.
(b) in reality, if an odd number multiplies and odd number, the result is also an odd number. Therefore, the question is wrong. I assume they wanted to ask for the proof that the product is also odd. If that's the case, then this is the proof:
Given that x and y are odd, we want to prove that xy is odd. For x and y to be odd, they have to be multiples of 2 with 1 added to it. Therefore, suppose x = 2k + 1 and y = 2p + 1 then xy = (2k + 1)(2p + 1) = 4kp + 2k + 2p + 1 = 2(kp + k + p) + 1. Let kp + k + p = q, then we have 2q + 1 which is also odd.
(c) Given that x is odd we want to prove that 3x is also odd. Firstly, we've proven above that xy is odd if x and y are odd. 3 is an odd number and we are told that x is odd. Therefore it follows from the second proof that 3x is also odd.
