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schepotkina [342]
3 years ago
12

Which is equivalent to cos^-1(.17) round your answer to the nearest hundredth of a radian

Mathematics
1 answer:
stellarik [79]3 years ago
4 0

Answer:

cos^-1(.17) is 1.40 in radian

Step-by-step explanation:

Given cos^-1(.17), the equivalent is 80.212

This is approximately 80.21°

Which is 1.40 in radian

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Without actually carrying out the steps of multiplication, determine whether the product of 15.1 and 0.9 will be larger or small
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Answer: Smaller

Step-by-step explanation: The product of 15.1 and 0.9 will be smaller than 15.1. This is because 0.9 is smaller than 1 and since 15.1 * 1 is 15.1, we know that anything smaller than 1 will result in the product of it and 15.1 being smaller.

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3 years ago
Given: KLMN is a trapezoid m∠N = m∠KML ME ⊥ KN , ME = 3 5 , KE = 8, LM KN = 3 5 Find: KM, LM, KN, Area of KLMN
prisoha [69]
A) Find KM∠KEM is a right angle hence ΔKEM is a right angled triangle Using Pythogoras' theorem where the square of hypotenuse is equal to the sum of the squares of the adjacent sides we can answer the 
KM² = KE² + ME²KM² = 8² + (3√5)²       = 64 + 9x5KM = √109KM = 10.44
b)Find LMThe ratio of LM:KN is 3:5 hence if we take the length of one unit as xlength of LM is 3xand the length of KN is 5x ∠K and ∠N are equal making it a isosceles trapezoid. A line from L that cuts KN perpendicularly at D makes KE = DN
KN = LM + 2x 2x = KE + DN2x = 8+8x = 8LM = 3x = 3*8 = 24
c)Find KN Since ∠K and ∠N are equal, when we take the 2 triangles KEM and LDN, they both have the same height ME = LD.
∠K = ∠N  Hence KE = DN the distance ED = LMhence KN = KE + ED + DN since ED = LM = 24and KE + DN = 16KN = 16 + 24 = 40
d)Find area KLMNArea of trapezium can be calculated using the  formula below Area = 1/2 x perpendicular height between parallel lines x (sum of the parallel sides)substituting values into the general equationArea = 1/2 * ME * (KN+ LM)          = 1/2 * 3√5 * (40 + 24)         = 1/2 * 3√5 * 64         = 3 x 2.23 * 32         = 214.66 units²
7 0
3 years ago
A carnival ride is in the shape of a wheel with a radius of 25 feet. The wheel has 20 cars attached to the center of the wheel.
Vladimir79 [104]

Centre angle will be 360° as its totally a circle.

  • radius=r=25ft

Angle between two cars=360/20=18°

  • Arc length=L

\\ \sf\longmapsto L=\dfrac{\theta}{360}(2\pi r)

\\ \sf\longmapsto L=\dfrac{18}{360}(2\pi(25))

\\ \sf\longmapsto L=\dfrac{1}{20}(50\pi)

\\ \sf\longmapsto L=2.5\pi ft

Now

Area be A

\\ \sf\longmapsto A=\dfrac{1}{2}Lr^2

\\ \sf\longmapsto A=\dfrac{1}{2}(2.5\pi)(25)^2

\\ \sf\longmapsto A=625(2.5)\pi\dfrac{1}{2}

\\ \sf\longmapsto A=1562.5\pi/2

\\ \sf\longmapsto A=781\pi ft^2=

4 0
3 years ago
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