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1 month ago

Which of the following equations correctly represents the law of sines?

1 answer:

5 0

The Law of Sines can be expressed as
b/sin(B) = c/sin(C)

Next, multiply both sides by sin(B).
This leads to b = (c*sin(B))/sin(C)

Thus, the correct response is C.

Answer: C. (c*sin B)/sin C

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An artist is creating a sculpture using bendable metal rods of equal length. One rod is formed into the shape of a square and an

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Refer to the image below. Note that the perimeters are identical since a single rod constitutes the perimeter, and since all rods are of equal length.

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1 month ago

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Explain why there are so many pennies on Rows 1-4. How do you think the number of pennies on Rows 5-8 will compare?

zzz [12365]

The variations occur due to the distinct rows of pennies

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1 month ago

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Find the point on the circle x^2+y^2 = 16900 which is closest to the interior point (30,40)

Leona [12618]

Response-

(78,104) represents the point closest to the interior.

Explanation-

The equation defining the circle,

$\Rightarrow x^2+y^2 = 16900$

$\Rightarrow y^2 = 16900-x^2$

$\Rightarrow y = \sqrt{16900-x^2}$

Since the point lies on the circle, its coordinates must be,

$(x,\sqrt{16900-x^2})$

The distance "d" from the point to (30,40) can be calculated as,

$=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$=\sqrt{(x-30)^2+(\sqrt{16900-x^2}-40)^2}$

$=\sqrt{x^2+900-60x+16900-x^2+1600-80\sqrt{16900-x^2}}$

$=\sqrt{9400-60x-80\sqrt{16900-x^2}}$

Next, we need to determine the value of x for which d is minimized. The minimum distance occurs when $9400-60x-80\sqrt{16900-x^2}$ is at its lowest value.

Let’s set up the equation,

$\Rightarrow f(x)=9400-60x-80\sqrt{16900-x^2}$

$\Rightarrow f'(x)=-60+80\dfrac{x}{\sqrt{16900-x^2}}$

$\Rightarrow f''(x)=\dfrac{1352000}{\left(16900-x^2\right)\sqrt{16900-x^2}}$

We find the critical points,

$\Rightarrow f'(x)=0$

$\Rightarrow-60+80\dfrac{x}{\sqrt{16900-x^2}}=0$

$\Rightarrow 80\dfrac{x}{\sqrt{16900-x^2}}=60$

$\Rightarrow 80x=60\sqrt{16900-x^2}$

$\Rightarrow 80^2x^2=60^2(16900-x^2)$

$\Rightarrow 6400x^2=3600(16900-x^2)$

$\Rightarrow \dfrac{16}{9}x^2=16900-x^2$

$\Rightarrow \dfrac{25}{9}x^2=16900$

$\Rightarrow x=\sqrt{\dfrac{16900\times 9}{25}}=78$

$\Rightarrow x=78$

Then,

$\Rightarrow f''(78)=\dfrac{1352000}{\left(16900-78^2\right)\sqrt{16900-78^2}}=\dfrac{125}{104}=1.2$

Since f''(x) is positive, the function f(x) achieves its minimum at x=78

When x is set to 78, the corresponding y value will be

$\Rightarrow y = \sqrt{16900-x^2}=\sqrt{16900-78^2}=104$

This leads us to conclude that the closest point is (78,104)

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2 months ago

Drag and drop an answer to each box to correctly complete the explanation for deriving the formula for the volume of a sphere.

lawyer [12517]

The formula for the volume of a sphere can be derived as follows. We will approach this through calculus, utilizing the concept of a solid of revolution; this is a three-dimensional shape formed by rotating a two-dimensional curve around a straight line (the axis of revolution) that lies within the same plane. From calculus, we know that we will generate a shape by rotating the specified circumference. Next, we isolate y and utilize certain limits for this integral.

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2 months ago

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Let P be a point not on the line L that passes through the points Q and R. The distance d from the point P to the line L is d =

AnnZ [12381]

Answer:

The distance from the point (0,1,1) to the specified line is zero.

Step-by-step explanation:

Considering the parametric equations of the line,

x=2t, y=5-2t, z=1+t

In order to calculate the distance from (0,1,1), we must remove t from the equations above, such that

$y=5-2t=5-x\implies x+y=5=4+1=4+z-t=4+z-\frac{1}{2}x$

$\implies 3x+2y-2z-8=0\hfill (1)$

whose direction ratios are (l,m,n)=(3,2,-2) and the distance from point (a,b,c)=(0,1,1) is defined as

$\frac{al+bm+cn}{\sqrt{l62+m^2+n^2}}=\frac{(3\times 0)+(2\times 1)+(-2)(1)}{\sqrt{3^2+2^2+(-2)^2}}=\frac{0+2-2}{\sqrt{17}}=0$

The distance between the point (0,1,1) and (1) amounts to zero. Therefore, the point (0,1,1) is located on the line (1).

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2 months ago