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Hunter-Best [27]
2 years ago
7

In the diagram, point O is the center of the circle and m∠ADB = 43°. If m∠AOB = m∠BOC, what is m∠BDC?

Mathematics
2 answers:
mr_godi [17]2 years ago
8 0

Answer:

B. 43

Step-by-step explanation:

I just took the test on plato :)

Sladkaya [172]2 years ago
7 0

Answer:

m∠BDC = 43°

Step-by-step explanation:

According to the theorem every peripheral angle in the circle is equal to half value of central angle.

Angle m∠ADB is corresponding peripheral angle of central angle m∠AOB.

According to this m∠AOB = 2· m∠ADB = 2· 43 = 86°

If angle m∠BOC=m∠AOB= 86°

Angle m∠BDC is corresponding peripheral angle of central angle m∠BOC

According to this m∠BDC = m∠BOC/2 = 43°

Good luck!!!

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Find the distance between P1(4,16degrees) and P2(-2,177degrees) on the polar plane.
bazaltina [42]
Polar coordinates give the distance from the origin and the angle from the positive x axis. Cartesian coordinates give the distance from the x and y axes.

You can draw a right triangle with these values. (see attached)
If you know the r value and theta of that triangle below, you can use trig to find x and y.

Let's convert (4, 16°) to Cartesian coordinates.

Note that since our angle is acute, (in Quadrant I) our sine and cosine will both be positive, as you should be able to derive from the unit circle, where cosine is represented as an x value and sine is represented as a y value.

cosine = adjacent / hypotenuse
cosθ = x/r
cos(16°) = x/4
4cos(16°) = x ≈ 3.84504678375

sine = oppsite / hypotenuse
sinθ = y/r
sin(16°) = y/4
4sin(16°) = y ≈ 1.10254942327<span>

So (4, 16°) </span>⇒ (3.84504678375, 1.10254942327).

Let's convert (-2, 177°)  to Cartesian coordinates.
Whenever you have a negative radius, that means to put the point opposite where it would have been if it had a positive radius. (see attached)

In that case, we can essentially add 180° to our current 177° to the same effect. That means that (-2, 177°) = (2, 357°).

Note that since our angle is in Quadrant IV, our cosine will be positive, but our sine will be negative. (as derived from the unit circle) We don't have to worry about this since our calculator figures this for us, but you should pay attention to it if you are converting from Cartesian to polar.

cosine = adjacent / hypotenuse
cosθ = x/r
cos(357°) = x/2
2cos(357°) = x ≈ 1.99725906951

sine = opposite / hypotenuse
sinθ = y/r
sin(357°) = y/2
2sin(357°) = y ≈ -0.10467191248

So (-2, 177°) ⇒ (1.99725906951, -0.10467191248).

Now we must use the distance formula with our two points.
d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
d\approx\sqrt{(1.99725906951-3.84504678375)^2+(-0.10467191248-1.10254942327)^2}
d\approx\sqrt{-1.84778771^2+-1.20722134^2}
d\approx\sqrt{3.41431942+1.45738336}
d\approx\sqrt{4.87170278}
\boxed{d\approx2.20719342}

7 0
2 years ago
Someone please help me!
Ludmilka [50]
The answer is 180 miles
4 0
2 years ago
Find the solution to this system of equations <br> 3x+2y+3z=3 4x-5y+7z=1 2x+3y-2z=6 <br> x=? Y=? Z=?
Ede4ka [16]

Answer:

The solution is x=2,\ y=0,\ z=-1.

Step-by-step explanation:

You are given the system of three equations:

\left\{\begin{array}{l}3x+2y+3z=3\\4x-5y+7z=1\\2x+3y-2z=6\end{array}\right.

Multiply the first equation by 4, the second equation by 3 and subtract them. Then multiply the third equation by 2 and subtract it from the second equation:

\left\{\begin{array}{l}3x+2y+3z=3\\4(3x+2y+3z)-3(4x-5y+7z)=4\cdot 3-3\cdot 1\\4x-5y+7z-2(2x+3y-2z)=1-2\cdot 6\end{array}\right.\Rightarrow \\\\\left\{\begin{array}{l}3x+2y+3z=3\\12x+8y+12z-12x+15y-21z=12-3\\4x-5y+7z-4x-6y+4z=1-12\end{array}\right.

So,

\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\-11y+11z=-11\end{array}\right.\Rightarrow \left\{\begin{array}{l}3x+2y+3z=3\\23y-9z=9\\y-z=1\end{array}\right.

Multiply the third equation by 23 and subtract it from the second equation:

\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23(y-z)=9-23\cdot 1\end{array}\right.\Rightarrow \left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23y+23z=9-23 \end{array}\right.

Hence,

\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\14z=-14 \end{array}\right.\Rightarrow z=-1

Substitute it into the second equation:

23y-9\cdot (-1)=9\Rightarrow 23y+9=9\\ \\23y=0\\ \\y=0

Substitute them into the first equation:

3x+2\cdot 0+3\cdot (-1)=3\Rightarrow 3x-3=3\\ \\3x=6\\ \\x=2

The solution is x=2,\ y=0,\ z=-1.

3 0
3 years ago
Please help please please ASAP please please
Lelu [443]

Step-by-step explanation:

ac is the answer

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4 0
3 years ago
How do you solve 3 3/4a+3+6=87 in multi step equations
Dovator [93]
If you would like to solve the equation 3 3/4 * a + 3 + 6 = 87, you can do it using the following steps:

3 3/4 * a + 3 + 6 = 87
15/4 * a = 87 - 3 - 6
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a = 78 * 4/15
a = 104/5
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The correct result is 20.8.
3 0
3 years ago
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