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Crank
3 years ago
5

Consider the following expression.

Mathematics
1 answer:
den301095 [7]3 years ago
6 0

Answer:

2 is a constant

3y + 5x + 2 is written as a sum of three terms

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3.5 squared. 12.25 times more would be the answer.
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What is the exact volume of the cylinder enter your answers in the term of pie in the Box 5cm 15cm​
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Valume of cylinder : πr² × h = π × 5² × 15 = 375π ≈ 1177.5 cm³

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A basketball team wants to paint half of a free-throw circle grey. If the circumference of the free-throw circle is 30.77 feet,
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Answer:

37.69    (≈ 38 ft²)

Step-by-step explanation:

2πr = 30.77

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What is the domain and range of the step function with the equation below? <br> f(x) = -3[ x ]
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Answer:

C

Step-by-step explanation:

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7 0
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Given that tan θ ≈ −0.087, where 3 2 π &lt; θ &lt; 2 , π find the values of sin θ and cos θ.
elena-s [515]

Answer:

  • sin θ ≈ -0.08667
  • cos θ ≈ 0.99624

Step-by-step explanation:

Straightforward use of the inverse tangent function of a calculator will tell you θ ≈ -0.08678 radians. This is an angle in the 4th quadrant, where your restriction on θ places it. (To comply with the restriction, you would need to consider the angle value to be 2π-0.08678 radians. The trig values for this angle are the same as the trig values for -0.08678 radians.)

Likewise, straightforward use of the calculator to find the other function values gives ...

  sin(-0.08678 radians) ≈ -0.08667

  cos(-0.08678 radians) ≈ 0.99624

_____

<em>Note on inverse tangent</em>

Depending on the mode setting of your calculator, the arctan or tan⁻¹ function may give you a value in degrees, not radians. That doesn't matter for this problem. sin(arctan(-0.087)) is the same whether the angle is degrees or radians, as long as you don't change the mode in the middle of the computation.

We have shown radians in the above answer because the restriction on the angle is written in terms of radians.

_____

<em>Alternate solution</em>

The relationship between tan and sin and cos in the 4th quadrant is ...

  \cos{\theta}=\dfrac{1}{\sqrt{1+\tan^2{\theta}}}\\\\\sin{\theta}=\dfrac{\tan{\theta}}{\sqrt{1+\tan^2{\theta}}}

That is, the cosine is positive, and the sign of the sine matches that of the tangent.

This more complicated computation gives the same result as above.

4 0
3 years ago
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