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Vinvika [58]
2 years ago
9

Find a (Round to the nearest tenth). PLS HURRY!!

Mathematics
2 answers:
nydimaria [60]2 years ago
6 0

Answer:

a = 56.3°

Step-by-step explanation:

tan(a) = 9/6 = 1.5

atan(1.5) = 56.31°

Gnom [1K]2 years ago
4 0

Answer:

\boxed {\boxed {\sf 56.3 \textdegree}}

Step-by-step explanation:

We are asked to find the measure of angle a.

This triangle is a right triangle because of the small triangle in the corner representing a 90 degree or right angle. Therefore, we can use trigonometric functions. The three main functions are:

  • sinθ=opposite/hypotenuse
  • cosθ= adjacent/hypotenuse
  • tanθ= opposite/adjacent

The side measuring 6 is adjacent or next to angle a. The side measuring 9 is opposite angle a. Therefore, we will use the tangent function.

tan \theta= \frac{ opposite}{adjacent}

tan  \ a = \frac{ 9}{6}

Since we are solving for an angle measure, we use the inverse trigonometric function.

tan ^{-1} * tan \ a = tan ^{-1} * \frac{9}6}

a= tan ^{-1} * \frac{9}6}

a= 56.30993247

Round to the nearest tenth. The 0 in the hundredth place tells us to leave the 3 in the tenth place.

a \approx 56.3

The measure of angle a is approximately 56.3 degrees.

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Answer:

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Step-by-step explanation:

To write 4.16666666 as a fraction you have to write 4.16666666 as numerator and put 1 as the denominator. Now you multiply numerator and denominator by 10 as long as you get in numerator the whole number.

5 0
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1. Evaluate the following:<br> a) 249 =<br> b) - 100<br> c) 10 + 7169
mario62 [17]

Answer:

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b) - 100 = - (10)²

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Step-by-step explanation:

3 0
3 years ago
Show that the following statement is an identity by transforming the left side into the right side. sin θ (sec θ + csc θ) = tan
ExtremeBDS [4]

Answer:

Step-by-step explanation:

Required to prove that:

Sin θ(Sec θ + Cosec θ)= tan θ+1

Steps:

Recall sec θ= 1/cos θ and cosec θ=1/sin θ

Substitution into the Left Hand Side gives:

Sin θ(Sec θ + Cosec θ)

= Sin θ(1/cos θ  + 1/sinθ )

Expanding the Brackets

=sinθ/cos θ + sinθ/sinθ

=tanθ+1 which is the Right Hand Side as required.

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8 0
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Find the area lying outside r=6cos(theta) and inside r=3+3cos(theta)
Sloan [31]
The area bounded by the 2 parabolas is A(θ) = 1/2∫(r₂²- r₁²).dθ between limits θ = a,b... 

<span>the limits are solution to 3cosθ = 1+cosθ the points of intersection of curves. </span>
<span>2cosθ = 1 => θ = ±π/3 </span>

<span>A(θ) = 1/2∫(r₂²- r₁²).dθ = 1/2∫(3cosθ)² - (1+cosθ)².dθ </span>
<span>= 1/2∫(3cosθ)².dθ - 1/2∫(1+cosθ)².dθ </span>
<span>= 9/8[2θ + sin(2θ)] - 1/8[6θ + 8sinθ +sin(2θ)] .. </span>
<span>.............where I have used ∫(cosθ)².dθ=1/4[2θ + sin(2θ)] </span>
<span>= 3θ/2 +sin(2θ) - sin(θ) </span>

<span>Area = A(π/3) - A(-π/3) </span>
<span>= 3π/6 + sin(2π/3) -sin(π/3) - (-3π/6) - sin(-2π/3) + sin(-π/3) </span>
<span>= π.</span>
7 0
3 years ago
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