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2 years ago

Please Help me To solve this Problem

Mathematics

1 answer:

mart [117]2 years ago

3 0

You're given that φ is an angle that terminates in the third quadrant (III). This means that both cos(φ) and sin(φ), and thus sec(φ) and csc(φ), are negative.

Recall the Pythagorean identity,

cos²(φ) + sin²(φ) = 1

Multiply the equation uniformly by 1/cos²(φ),

cos²(φ)/cos²(φ) + sin²(φ)/cos²(φ) = 1/cos²(φ)

1 + tan²(φ) = sec²(φ)

Solve for sec(φ) :

sec(φ) = - √(1 + tan²(φ))

Given that cot(φ) = 1/4, we have tan(φ) = 1/cot(φ) = 1/(1/4) = 4. Then

sec(φ) = - √(1 + 4²) = -√17

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The rule is bring up the lower fraction and flip it to multiply, so that will give us:

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3 years ago

Evaluate √x^2 − y^2 when x^2 =33 and y^2 =15

Andre45 [30]

Answer:

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Step-by-step explanation:

In this question, we are asked to get the value of √ ( x²-y² ), while the value of x² and y² are given to be 33 and 15 respectively.

Hence, x² =33 and y² = 15.

Putting the values of x² and y² in the above expression we get,

√ ( x²-y² ) = √ ( 33-15) = √18 =3√2. (Answer)

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2 years ago

Can anyone help me?

Charra [1.4K]

  V = lwh
  2x³ + 17x² + 46x + 40 = l(x + 4)(x + 2)
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2x(x²) + 2x(6x) + 2x(8) + 5(x²) + 5(6x) + 5(8) = l(x + 4)(x + 2)
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4 0

3 years ago

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