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gulaghasi [49]
3 years ago
15

Given: tangent to Circle O. If m C = 57°, then m BDR =

Mathematics
1 answer:
Romashka-Z-Leto [24]3 years ago
8 0
∠BCD = 57°
∴ ∠BDR = ∠BCD = 57° (angle that meets the chord and the tangent is equi-angular to the angle at the alternate segment)
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6) Two coastguard stations P and Q are 17km apart, with due East of P. A ship S is observed in distress on a bearing 048° fromP
adelina 88 [10]

Answer:

The ship S is at 10.05 km to coastguard P, and 12.70 km to coastguard Q.

Step-by-step explanation:

Let the distance of the ship to coastguard P be represented by x, and its distance to coastguard Q be represented by y.

But,

<P = 048°

<Q = 360^{o} - 324^{o}

     = 036^{o}

Sum of angles in a triangle = 180^{o}

<P + <Q + <S = 180^{o}

048° + 036^{o} + <S = 180^{o}

84^{o} + <S = 180^{o}

<S  = 180^{o} -  84^{o}

    = 96^{o}

<S = 96^{o}

Applying the Sine rule,

\frac{y}{Sin P} = \frac{x}{Sin Q} = \frac{z}{Sin S}

\frac{y}{Sin P} = \frac{z}{Sin S}

\frac{y}{Sin 48^{o} } = \frac{17}{Sin 96^{o} }

\frac{y}{0.74314} = \frac{17}{0.99452}

⇒ y = \frac{12.63338}{0.99452}

       = 12.703

y = 12.70 km

\frac{x}{Sin Q} = \frac{z}{Sin S}

\frac{x}{Sin 36^{o} } = \frac{17}{Sin 96^{o} }

\frac{x}{0.58779} = \frac{17}{0.99452}

⇒ x = \frac{9.992430}{0.99452}

      = 10.0475

x = 10.05 km

Thus,

the ship S is at a distance of 10.05 km to coastguard P, and 12.70 km to coastguard Q.

6 0
3 years ago
Please help me thanks please <br> Will give you brainlest
Akimi4 [234]

Answer:

224

Step-by-step explanation:

<u>Base:</u>

A = Bh

A = 8(8) = 64

<u>Sides:</u>

A = 1/2 Bh

A = 1/2 8(10)

A = 40

<u>Then you do 40(4) because there are 4 sides:</u>

40 (4) = 160

<u>Then you add up the base and the sides:</u>

160 + 64 = 224

(I think this is the right answer)

7 0
3 years ago
A student desk measures 1.5 meters. The dry erase board measures 3.5 meters. How much longer is the dry erase board than the des
Serga [27]
The dry erase board is 200 centimeters longer.
4 0
3 years ago
Read 2 more answers
PLS HELP THIS IS HARD ANYONE PLS
Valentin [98]

This question is very oddly worded.  The domain is the set of x-values, but this is a set of (x,y) ordered pairs.

I'm reading this question as "Here's a function, { (1,5), (2,1), (-1,-7) }.  If this is reflected over the x-axis, what's the range?"

Assuming that is the question that is meant to be asked, reflecting a function over the x-axis will just change the signs of the y-values.

    (1,5) -> (1,–5)

    (2,1) -> (2,–1)

    (-1,-7) -> (-1,+7)

I'd pick the third option.

6 0
3 years ago
Finding Derivatives Implicity In Exercise, find dy/dx implicity.<br> 4x3 + ln y2 + 2y = 2x
lisov135 [29]

Answer:

\frac{dy}{dx}=\frac{y(1-6x^2)}{(1+y)}

Step-by-step explanation:

Given function : 4x³ + ln(y²) + 2y = 2x

on differentiating both sides with respect to 'x', we get

\frac{d(4x^3 + ln(y^2) + 2y)}{dx}= \frac{d(2x)}{dx}

or

12x^2+\frac{2}{y}(\frac{dy}{dx})+2(\frac{dy}{dx})=2

or

12x^2+(\frac{2}{y}+2)\frac{dy}{dx}=2

or

(\frac{2+2y}{y})\frac{dy}{dx}=2-12x^2

or

\frac{dy}{dx}=\frac{y(2-12x^2)}{2+2y}

or

\frac{dy}{dx}=\frac{2y(1-6x^2)}{2(1+y)}

or

\frac{dy}{dx}=\frac{y(1-6x^2)}{(1+y)}

3 0
3 years ago
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