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3 years ago

I’m looking for what kind of angles 2 and 6 are?

Mathematics

1 answer:

schepotkina [342]3 years ago

6 0

Answer:

Angles 2 and 11 are alternate interior angles I believe

Hope this helped!

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Which of the following functions have a vertical asymptote for values of Ø such that cos Ø = 1?Select all that apply.

igomit [66]

Answer:

$y=\cot \theta$ and $y=\csc \theta$

Step-by-step explanation:

Note that if $\cos \theta=1,$ then $\sin \theta=0.$

Functions $y=\cos \theta,\ \ y=\sin \theta$ do not have vertical asymptotes at all.

Vertical asymptotes have functions $y=\tan \theta,\ \ y=\cot \theta,\ \ y=\sec \theta,\ \ y=\csc \theta.$

Functions $y=\tan \theta$ and $y=\sec \theta$ have the same vertical asymptotes (when $\cos \theta =0$ ).

Functions $y=\cot \theta$ and $y=\csc \theta$ have the same vertical asymptotes (when $\sin \theta =0$ ). See attached diagram

3 0

3 years ago

Caleb needs at least 30 cups for the party he already has 12 cups at least show many more cups does he need

Maurinko [17]

B. Equation (hope this helps!)

5 0

3 years ago

URGENT I NEED HELP!!!!!!!!!!!!!!!!!!!!

Nataliya [291]

Answer:

$11

Step-by-step explanation:

66/6=11

3 0

3 years ago

The angles of depression of two ships from the top of the light house are 45° and 30° towards east. If the ships are 200 m apart

azamat

Answer:

273 meters

Step-by-step explanation:

See image attached for the diagram I used to represent this scenario.

The distance between the ships, at angles 30 and 45, is 200 meters. The distance between the left ship and the lighthouse is x meters.

We can use trigonometric ratios to solve this problem. We can use the tangent ratio $\big{(} \frac{\text{opposite}}{\text{adjacent}} \big{)}$ to create an equation with the two angles.

$\displaystyle \text{tan(45)} = \frac{h}{x}$
$\displaystyle \text{tan(30)} = \frac{h}{x+200} }$

Let's take these two equations and solve for x in both of them.

<h2> $\textbf{Equation I}$ </h2>

$\displaystyle \text{tan(45)} = \frac{h}{x}$

tan(45) = 1, so we can rewrite this equation.

$\displaystyle 1=\frac{h}{x}$

Multiply x to both sides of the equation.

$\displaystyle x = h$

<h2> $\textbf{Equation II}$ </h2>

$\displaystyle \text{tan(30)} = \frac{h}{x+200} }$

Multiply x + 200 to both sides and divide h by tan(30).

$\displaystyle \text{x + 200} = \frac{h}{\text{tan (30)}}$

Subtract 200 from both sides of the equation.

$\displaystyle \text{x} = \frac{h}{\text{tan (30)}} - 200$

Simplify h/tan(30).

$x=\sqrt{3}h - 200$

<h2> $\textbf{Equation I = Equation II}$ </h2>

Take Equation I and Equation II and set them equal to each other.

$h=\sqrt{3}h-200$

Subtract √3 h from both sides of the equation.

$h-\sqrt{3}h=-200$

Factor h from the left side of the equation.

$h(1-\sqrt{3}) =-200$

Divide both sides of the equation by 1 - √3.

$\displaystyle h=\frac{-200}{1-\sqrt{3} }$

Rationalize the denominator by multiplying the numerator and denominator by the conjugate.

$\displaystyle h=\frac{-200}{1-\sqrt{3} } \big{(} \frac{1+\sqrt{3} }{1+\sqrt{3}} \big{)}$
$\displaystyle h=\frac{-200+200\sqrt{3} }{1-3}$
$\displaystyle h =\frac{-200+200\sqrt{3} }{-2}$