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2 years ago

Help me pleae HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

Mathematics

1 answer:

max2010maxim [7]2 years ago

3 0

Answer:

Step-by-step explanation:

d = c/pi

so d = 62.8/pi

d = 62.8/pi or appr 19.99in

r = d/2

so r = 62.8/2pi or 31.4/pi appr 19.99/2

$a = \pi r^{2}$

$a = \pi *(31.4/\pi)^2$

then

$a = 985.96/pi$

a = approximately 313.84

answer is rowan

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What is 3/4 / 1/6? (please help)

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3 years ago

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I need help with this problem. Given: BC = 10 inches AC = √50 inches m∠CBD = 60° m∠CAD = 90° Calculate the exact area of the sha

Nikitich [7]

Answer:

Area of the shaded region = 23.33 in²

Step-by-step explanation:

Area of a sector = $\frac{\theta}{360}(\pi r^{2})$

Where θ = Central angle subtended by an arc

r = radius of the circle

Area of the sector BCD = $\frac{60}{360}(\pi) (10^{2})$

= 52.36 in²

Area of equilateral triangle BCD = $\frac{\sqrt{3} }{4}(\text{Side})^2$

= $\frac{\sqrt{3} }{4}(10)^2$

= $25\sqrt{3}$ in²

= 43.30 in²

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= 9.06 in²

Area of sector CAD = $\frac{90}{360}(\pi)(\sqrt{50})^2$

= 39.27 in²

Area of right triangle CAD = $\frac{1}{2}(\text{Base})(\text{Height})$

= $\frac{1}{2}(\text{AC})(\text{AD})$

= $\frac{1}{2}(\sqrt{50})(\sqrt{50})$

= 25 in²

Area of the shaded part in the ΔACD = 39.27 - 25

= 14.27 in²

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8 0

3 years ago

HELP Use either law of sines or law of cosine. Need help on this problem! show work please!

Marina86 [1]

Answer: x = 15.035677095729 approximately

Round this however you need to.

=================================================

Explanation:

I'm assuming you want to find the value of x, which your diagram is showing to be the length of segment QR.

If so, then we'll need to find the measure of angle Q first. Using the law of sines, we get the following:

sin(Q)/q = sin(R)/r

sin(Q)/PR = sin(R)/PQ

sin(Q)/13 = sin(85)/19

sin(Q) = 13*sin(85)/19

sin(Q) = 0.6816068987

Q = arcsin(0.6816068987) ... or ... Q = 180-arcsin(0.6816068987)

Q = 42.9693397461 ... or ... Q = 137.0306602539

These values are approximate.

----------------

Now if Q = 42.9693397461 approximately, then angle P is

P = 180-Q-R

P = 180-42.9693397461-85

P = 52.0306602539

Similarly, if Q = 137.0306602539 approximately, then,

P = 180-Q-R

P = 180-137.0306602539-85

P = -42.0306602539

A negative angle is not possible, so we'll ignore Q = 137.0306602539

----------------

The only possible value of angle P is approximately P = 52.0306602539

Let's apply the law of sines again to find side p, aka segment QR

sin(P)/p = sin(R)/r

sin(P)/QR = sin(R)/PQ

sin(52.0306602539)/x = sin(85)/19

19*sin(52.0306602539) = x*sin(85)

19*sin(52.0306602539)/sin(85) = x

x = 15.035677095729

This value is approximate.

Round this value however you need to.

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3 years ago

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Triangle PQR is transformed to triangle P'Q'R'. Triangle PQR has vertices P(4, 0), Q(0, −4), and R(−8, −4). Triangle P'Q'R' has

Whitepunk [10]

Answer:

Please find attached the required plot accomplished with an online tool

Part A:

1/4

Part B:

P''(-1, 0), Q''(0, -1), and R''(2, -1)

Part C:

Triangle PQR is similar to triangle P''Q''R'' but they are not congruent

Step-by-step explanation:

Part A:

Triangle ΔPQR has vertices P(4, 0), Q(0, -4), R(-8, -4)

Triangle ΔP'Q'R' has vertices P'(1, 0), Q'(0, -1), R'(-2, -1)

The dimensions of the sides of the triangle are given by the relation;

$l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}$

Where;

(x₁, y₁) and (x₂, y₂) are the coordinates on the ends of the segment

For segment PQ, we place (x₁, y₁) = (4, 0) and (x₂, y₂) = (0, -4);

By substitution into the length equation, we get;

The length of segment PQ = 4·√2

The length of segment PR = 4·√10

The length of segment RQ = 8

The length of segment P'Q' = √2

The length of segment P'R' = √10

The length of segment R'Q' = 2

Therefore, the scale factor of the dilation of ΔPQR to ΔP'Q'R' is 1/4

Part B:

Reflection of (x, y) across the y-axis gives;

(x, y) image after reflection across the y-axis = (-x, y)

The coordinates after reflection of P'(1, 0), Q'(0, -1), R'(-2, -1) across the y-axis is given as follows;

P'(1, 0) image after reflection across the y-axis = P''(-1, 0)

Q'(0, -1) image after reflection across the y-axis = Q''(0, -1)

R'(-2, -1) image after reflection across the y-axis = R''(2, -1)

Part C:

Triangle PQR is similar to triangle P''Q''R'' but they are not congruent as the dimensions of the sides of triangle PQR and P''Q''R'' are not the same.

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3 years ago