What figure is always a rectangle?

Let theta be an angle in quadrant II such that cos theta = -2/3

Iteru [2.4K]

Answer:

So we have $\csc(\theta)=\frac{3 \sqrt{5}}{5} \text{ and } \tan(\theta)=\frac{-\sqrt{5}}{2}$ .

Step-by-step explanation:

Ok so we are in quadrant 2, that means sine is positive while cosine is negative.

We are given $\cos(\theta)=\frac{-2}{3}(\frac{\text{adjacent}}{\text{hypotenuse}})$ .

So to find the opposite we will just use the Pythagorean Theorem.

$a^2+b^2=c^2$

$(2)^2+b^2=(3)^2$

$4+b^2=9$

$b^2=5$

$b=\sqrt{5}$ This is the opposite side.

Now to find $\csc(\theta)$ and $\tan(\theta)$ .

$\csc(\theta)=\frac{\text{hypotenuse}}{\text{opposite}}=\frac{3}{\sqrt{5}}$ .

Some teachers do not like the radical on bottom so we will rationalize the denominator by multiplying the numerator and denominator by sqrt(5).

So $\csc(\theta)=\frac{3}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}=\frac{3 \sqrt{5}}{5}$ .

And now $\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}=\frac{\sqrt{5}}{-2}=\frac{-\sqrt{5}}{2}$ .

So we have $\csc(\theta)=\frac{3 \sqrt{5}}{5} \text{ and } \tan(\theta)=\frac{-\sqrt{5}}{2}$ .

7 0

3 years ago

Read 2 more answers

Find the volume of a cylinder that has a radius of 12 mm

Mrac [35]

Answer:

8,930.16

Step-by-step explanation:

the equation for the volume of a cylinder is 3.14(pie) x the radius squared x the height of the cylinder.

8 0

3 years ago

Directions: Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

Semmy [17]

9 pens and 15 puzzles.....GCF is 3
9/3 = 3 : 15/3 = 5
he can make 3 bags, each containing 3 pens and 5 puzzles

6 soft candies and 18 pencils....GCF = 6
6/6 = 1 : 18/6 = 3
he can make 6 bags, each containing 1 soft candy and 3 pencils

8 chocolate coins and 12 hard candies....GCF = 4
8/4 = 2 : 12/4 = 3
he can make 4 bags, each containing 2 chocolate coins and 3 hard candies

8 0

3 years ago

Which numbers are solutions of the inequality 3x – 4 > 5?

Lyrx [107]

Answer:

X>3

Step-by-step explanation:

Inequality form:

x>3

Hope this helped there wasn't much information sorry

4 0

3 years ago

Find all subsets of the set s = {(i, 0), (0, i), (i, i)} that form a basis for r2•

sasho [114]

All subsets of the set s = {(i, 0), (0, i), (i, i)} that form a basis for r2 is

{( 0,1) (-1,1)}

{(1,0)(0,1)}

{(1,0)(-1,1)}

A fixed A is a subset of some other set B if all elements of set A are factors of set B. In other phrases, set A is contained within set B. The subset relationship is denoted as A⊂B.

Subsets are a part of one of the mathematical ideas known assets. a hard and fast is a set of items or factors, grouped within the curly braces, consisting of {a,b,c,d}. If a set A is a set of even numbers and set B includes {2, 4,6}, then B is stated to be a subset of A, denoted by B⊆A and A is the superset of B.

In mathematics, set A is a subset of a fixed B if all factors of A are also elements of B; B is then a superset of A. It's far possible for A and B to be equal; if they may be unequal, then A is a proper subset of B. The relationship of one set being a subset of every other is known as inclusion.

A set which is (is linearly independent andoil generates/ spans the space (like R) is called a baris.

S= (1,0), (0,1), (-1))}

option-1

As, dimension of R²=2, so, Ary set

conists of three vectors cannot be baris for R²

so, option-1 is wrong.

option-2

As, dimention of R2-2.

{(10), (-11)} is a linearly independent

subset of S. so And as, this subset

has 2 vectors so it will as also

span R se, it will be a baris

of IR?

So, option-6 is correct

{(0,1)} is linearly independent subset

option-7

As (LO) has one vector. So, it cannot

Span/generate R so, it also cannot be balis

for R2

of S.

But it cannot span R²

So ut cannot

be a baris of R2

se, option-2 is wrong.

Option-3

AS, {(0,1), (-1,1)} is linearly independent

So, option -7, is wrong.

Answer

[as fox amy KER, (-1,1) K (0,1)] subset of s

And also as dim {(0,1), (-11)}=2

which is same as the dimention of IR2

so {(0,1), (-11)} forms a baris of R2

{(1,0), (0,1), (-11)}

{(1,0)}

So, option-3 is

Connect

option-4

{(1,0), (3, 1)} is standard balis of R2

which is subset of S.

10

[{(1,0), (0,1)}]

So, option-4 is correct.

option-5

{(-11)}

consists of one vertex,

so, ut cannot span R², so, ut cannot

be a baris of R²

9090

LO {(1,0), (-1,1)}]

Learn more about subsets here brainly.com/question/2000547

#SPJ4

7 0

2 years ago