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2 years ago

Is sqrt(1 - sin^2 θ) = cos θ true? if so, in which quadrants does angle θ terminate?

Mathematics

1 answer:

jonny [76]2 years ago

7 0

Answer:

I and IV

Step-by-step explanation:

Since 1-sin(θ)² = cos(θ)², the given equation is equivalent to ...

√(cos(θ)²) = |cos(θ)| = cos(θ)

This will be true where the cosine is non-negative, in the first and fourth quadrants.

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The length of a rectangle is represented by the polynomial 2x3 – 5x2 + 8 and the width is represented

Rom4ik [11]

1) the polynomial representing the area is

2x^4 + 6x^3 - 5x^2 -7x + 24

2) the constant term is

3) the polynomial is a 4th degree

4) if my guess is ryt... the LEADING coefficient is the coefficient of the highest degree of X ...i.e 2

8 0

2 years ago

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Please help math middle school level

pychu [463]

Answer:

7 yd

Step-by-step explanation:

You are finding the perimeter. Your answer should include yd (yd² is for area, yd³ is for 3d images).

Note that the two lines segments (? & 9) is parallel to the other line segment (16). Find the ?:

? + 9 = 16

? + 9 (-9) = 16 (-9)

? = 16 - 9

? = 7

7 yd is your answer

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3 years ago

Factor 9(n − 4) − 4n(n − 4)

schepotkina [342]

Hey there :)

9( n - 4 ) - 4n ( n - 4 )
We can take ( n - 4 ) as common as both are getting multiplied by ( n - 4 )

We can join 9 - 4n together

Therefore, the final answer will be ( n - 4 )( 9 - 4n )

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2 years ago

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From the side view, a gymnastics mat forms a right triangle with other angles measuring 60° and 30°. The gymnastics mat extends

valkas [14]

Answer: The mat is 4.33 ft high off the ground.

Step-by-step explanation:

Since we have given that

Angle of elevation with the first triangle = 30°

Angle of elevation with the second triangle = 60°

Length at which gymnastics mat extends across the floor = 5 feet

so, As shown in the figure:

We need to find the height of the mat off the ground.

If CD = 5 ft,

Let, AB = y, DC = x.

In Δ ABC,

$\tan 60^\circ=\frac{AB}{BC}\\\\\frac{\sqrt{3}}{2}=\frac{y}{x}\\\\x=\frac{y}{\sqrt{3}}$

Similarly, in Δ ACD,

$\tan 30^\circ=\frac{AB}{BD}\\\\\frac{1}{\sqrt{3}}=\frac{y}{x+5}\\\\\frac{1}{\sqrt{3}}=\frac{y}{\frac{y}{\sqrt{3}}+5}\\\\\frac{1}{\sqrt{3}}=\frac{y\sqrt{3}}{y+5\sqrt{3}}\\\\3y=y+5\sqrt{3}\\\\2y=5\sqrt{3}\\\\y=\frac{5\sqrt{3}}{2}\\\\y=4.33\ ft$