Ask question

Ask question

All categories

2 years ago

6

Can someone help me plz...I think the answer is 24 but I'm not sure.. thanks :)

2 answers:

Trava [24]2 years ago

7 0

24 is the correct answer your right

ad-work [718]2 years ago

7 0

Y = 12/5 x . You're correct.

You might be interested in

Classify Angle 1 & Angle 2 by choosing the correct term from the choices below.

valkas [14]

Answer:

Vertical

Step-by-step explanation:

Intersecting lines form Vertical angles.

6 0

1 year ago

HELP ASAP FIRST RIGHT GET BRAINLEST

My name is Ann [436]

Answer:

A

Step-by-step explanation: hope it helps

7 0

2 years ago

Read 2 more answers

43: PLEASE HELP Write the equation 22x-2y-10 in slope-intercept form

Gnoma [55]

Answer:

y= 11x -5

Step-by-step explanation:

I believe you rearrange it to get y on one side of the equation so,

2y= 22x-10

and then divide 2 on both sides to get,

y=11x-5

7 0

2 years ago

Find the value of X.

ch4aika [34]

Answer:

x= 60

Step-by-step explanation:

if y is the other missing side

y^2 = 64^2 + 48^2

y^2 = 4,096 + 2,304

y^2 = 6,400

100^2 = x^2 +y^2

x^2 = 10,000 - 6,400

x^2 = 3,600

x = 60

6 0

2 years ago

Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

4 0

2 years ago