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vivado [14]
3 years ago
6

A scale drawing of a rectangular basketball court measures 7 inches long and 4 inches wide. If the scale used for the drawing is

2.5 inches equals 5 yards, what is the actual length of the park?
Mathematics
1 answer:
gavmur [86]3 years ago
4 0

Answer:

14 yards long and 8 yards long

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Dilate point a in relation to the origin by a dilation factor of 5.?
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origin dilation if 5. is 5

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The sum of 5x and 2x is at least 14
tatuchka [14]

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4 years ago
Read 2 more answers
I need help with this
Flauer [41]
The given angles are complementary, therefore:

(5r + 5) + (8r -6) = 90°

13r - 1 = 90
13r - 1 + 1 = 90 + 1
13r = 91

13r/13 = 91/13

r = 7
6 0
3 years ago
Find the slope of the tangent line to the given polar curve at the point specified by the value of θ
MariettaO [177]

The given curve has equation

<em>r(θ)</em> = 9 + 8 cos(<em>θ</em>)

and its derivative is

d<em>r</em>/d<em>θ</em> = -8 sin(<em>θ</em>)

When <em>θ</em> = <em>π</em>/3, we have <em>r</em> (<em>π</em>/3) = 13, and d<em>r</em>/d<em>θ</em> (<em>π</em>/3) = -4√3.

Differentiate these with respect to <em>θ</em> :

d<em>y</em>/d<em>θ</em> = d<em>r</em>/d<em>θ</em> sin(<em>θ</em>) + <em>r(θ)</em> cos(<em>θ</em>)

d<em>x</em>/d<em>θ</em> = d<em>r</em>/d<em>θ</em> cos(<em>θ</em>) - <em>r(θ</em>) sin(<em>θ</em>)

In polar coordinates, we have

<em>y(θ)</em> = <em>r(θ)</em> sin(<em>θ</em>)

<em>x(θ)</em> = <em>r(θ)</em> cos(<em>θ</em>)

and when <em>θ</em> = <em>π</em>/3, we have <em>y</em> (<em>π</em>/3) = 13√3/2 and <em>x</em> (<em>π</em>/3) = 13/2.

The slope of the tangent line to the curve is d<em>y</em>/d<em>x</em>. By the chain rule,

d<em>y</em>/d<em>x</em> = d<em>y</em>/d<em>θ</em> • d<em>θ</em>/d<em>x</em> = (d<em>y</em>/d<em>θ</em>) / (d<em>x</em>/d<em>θ</em>)

d<em>y</em>/d<em>x</em> = (d<em>r</em>/d<em>θ</em> sin(<em>θ</em>) + <em>r(θ)</em> cos(<em>θ</em>)) / (d<em>r</em>/d<em>θ</em> cos(<em>θ</em>) - <em>r(θ</em>) sin(<em>θ</em>))

When <em>θ</em> = <em>π</em>/3, the slope is

d<em>y</em>/d<em>x</em> = (-4√3 sin(<em>π</em>/3) + 13 cos(<em>π</em>/3)) / (-4√3 cos(<em>π</em>/3) - 13 sin(<em>π</em>/3))

d<em>y</em>/d<em>x</em> = (-4√3 (√3/2) + 13 (1/2)) / (-4√3 (1/2) - 13 (√3/2))

d<em>y</em>/d<em>x</em> = - 1/(17√3)

So, the tangent line has slope -1/(17√3) and passes through (13/2, 13√3/2). Using the point-slope formula, its equation is

<em>y</em> - 13√3/2 = -1/(17√3) (<em>x</em> - 13/2)

<em>y</em> = -(<em>x</em> - 338)/(17√3)

4 0
3 years ago
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