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Bad White [126]

2 years ago

8

What is quadrant of (-3, -4)

2 answers:

guajiro [1.7K]2 years ago

7 0

Answer:

The point (-3,-4) would be in the lower left quadrant of a graph, also known as quadrant 3

Step-by-step explanation:

X and Y are both negative

ch4aika [34]2 years ago

7 0

Answer:

Third quadrant

Step-by-step explanation:

the Y- intercepts And the X-intercepts they are negative in the third quadrant

I hope you'll find these useful

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2.1e-10 in standard form?

just olya [345]

<span>2.1e-10
= 2.1 x 10^-10
= 0.00000000021

hope it helps</span>

5 0

3 years ago

Read 2 more answers

AQRS is a right triangle.

Paladinen [302]

The third one is correcte

5 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

3 years ago

Do you need to regroup 56+3 when adding to a two digit number

artcher [175]

You don't need to regroup when adding a two-digit number with a number.

3 0

3 years ago

WILL GIVE BRAINLIEST!!!

LUCKY_DIMON [66]

Answer:

A = 119 units²

Step-by-step explanation:

side of square is (12√2)sin45 = 12

The portion of the diagonal that is NOT octagon is

12√2 - 12

There are four corner triangles. The sum of their areas is

(12√2 - 12)²

Area of the octagon is area of the square minus four corner triangle areas.

A = 12² - (12√2 - 12)²

A = 119.2935...

4 0

3 years ago