The constant is the part of the algebraic function or expression that does not change. In this case, as x changes, the constant -10 does not. Thus, -10 is the constant.

4 0

3 years ago

If the radius of a circle is irrational, the area must be irrational

nirvana33 [79]

That's a true fact. If you want to get really technical, the area of a circle is. ALWAYS irrational because it's the product of something and PI.

8 0

3 years ago

Help, Help, Please

Lesechka [4]

Answer:

θ = $\frac{\pi }{3}$ (60° )

Step-by-step explanation:

Using the identity

sin²x + cos²x = 1 ⇒ sin²x = 1 - cos²x

Given

cos²θ - sin²θ = 2 - 5cosθ

cos²θ - (1 - cos²θ) = 2 - 5cosθ

cos²θ - 1 + cos²θ = 2 - 5cosθ

2cos²θ - 1 = 2 - 5cosθ ( subtract 2 - 5cosθ from both sides )

2cos²θ + 5cosθ - 3 = 0 ← in standard form

(cosθ + 3)(2cosθ - 1) = 0 ← in factored form

Equate each factor to zero and solve for θ

cosθ + 3 = 0

cosθ = - 3 ← not possible as - 1 ≤ cosθ ≤ 1

2cosθ - 1 = 0

cosθ = $\frac{1}{2}$ , so

θ = $cos^{-1}$ ( $\frac{1}{2}$ ) = $\frac{\pi }{3}$ ( or 60° )

8 0

3 years ago

Read 2 more answers

find the value of the trigonometric function sin (t) if sec t = -4/3 and the terminal side of angle t lies in quadrant II

kodGreya [7K]

Answer:

$sin(t) =\frac{\sqrt{7}}{4}$

Step-by-step explanation:

By definition we know that

$sec(t) = \frac{1}{cos(t)}$

and

$cos ^ 2(t) = 1-sin ^ 2(t)$

As $sec(t) = -\frac{4}{3}$

Then

$sec(t) = -\frac{4}{3}\\\\\frac{1}{cos(t)} =-\frac{4}{3}\\\\cos(t) = -\frac{3}{4}$

Now square both sides of the equation:

$cos^2(t) = (-\frac{3}{4})^2$

$cos^2(t) = \frac{9}{16}\\\\$

$1-sin^2(t) =\frac{9}{16}\\\\sin^2(t) =1-\frac{9}{16}\\\\sin^2(t) =\frac{7}{16}\\\\sin(t) =\±\sqrt{\frac{7}{16}}$

In the second quadrant sin (t) is positive. Then we take the positive root

$sin(t) =\sqrt{\frac{7}{16}}$

$sin(t) =\frac{\sqrt{7}}{4}$

8 0

3 years ago