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makkiz [27]
3 years ago
7

For the quadrant in which the following point is located, determine which of the functions are positive.

Mathematics
1 answer:
Kruka [31]3 years ago
6 0
Cot tan Ben. Skin tin
You might be interested in
I need help finding the missing side length with the answer in radicals in the simplest form.
AlekseyPX

Answer:

C

Step-by-step explanation:

There’s nothing hard with this

You need to know special right triangles

This is a 30,60, 90 triangle

We automatically know becuase theres. Right angle (90) and a 60 degree angle

now the formula is the smallest side is (n) {I’m using N becuase theres already an X) or in this case, 6

The hypotenuse, the one directly above the 2nd biggest side, or the diagnoal side is 2n or (12). Now we know what X is in this situation which is 12. So that narrows it to 2 answers

Now the side on the bottom of the hypotenuse, the second biggest side is used in the formula N\sqrt{3}. So we know what N is in the beginning, 6 so we just plug that in and well get C. Attached is a photo on q 30, 60, 90 special right triangle

4 0
3 years ago
A cube has a volume of 64cm cubed. What is the area of one face of the cube?
goldfiish [28.3K]

Answer:

96 cm squared

Step-by-step explanation:

5 0
3 years ago
Determine the gradient of the straight line 2x-3y+9=0. Find the equation of the straight line through the origin which is perpen
frutty [35]

Answer: 3

x

−

2

y

−

15

=

0

Explanation:

We know that,

the slope of the line  

a

x

+

b

y

+

c

=

0

is  

m

=

−

a

b

∴

The slope of the line  

2

x

+

3

y

=

9

is  

m

1

=

−

2

3

∴

The slope of the line perpendicular to  

2

x

+

3

y

=

9

is  

m

2

=

−

1

m

1

=

−

1

−

2

3

=

3

2

.

Hence,the equn.of line passing through  

(

3

,

−

3

)

and

m

2

=

3

2

is

y

−

(

−

3

)

=

3

2

(

x

−

3

)

y

+

3

=

3

2

(

x

−

3

)

⇒

2

y

+

6

=

3

x

−

9

⇒

3

x

−

2

y

−

15

=

0

Note:

The equn.of line passing through  

A

(

x

1

,

y

1

)

and

with slope

m

is

y

−

y

1

=

m

(

x

−

x

1

)3

x

−

2

y

−

15

=

0

Explanation:

We know that,

the slope of the line  

a

x

+

b

y

+

c

=

0

is  

m

=

−

a

b

∴

The slope of the line  

2

x

+

3

y

=

9

is  

m

1

=

−

2

3

∴

The slope of the line perpendicular to  

2

x

+

3

y

=

9

is  

m

2

=

−

1

m

1

=

−

1

−

2

3

=

3

2

.

Hence,the equn.of line passing through  

(

3

,

−

3

)

and

m

2

=

3

2

is

y

−

(

−

3

)

=

3

2

(

x

−

3

)

y

+

3

=

3

2

(

x

−

3

)

⇒

2

y

+

6

=

3

x

−

9

⇒

3

x

−

2

y

−

15

=

0

Note:

The equn.of line passing through  

A

(

x

1

,

y

1

)

and

with slope

m

is

y

−

y

1

=

m

(

x

−

Explanation:

the equation of a line in  

slope-intercept form

is.

∙

x

y

=

m

x

+

b

where m is the slope and b the y-intercept

rearrange  

2

x

+

3

y

=

9

into this form

⇒

3

y

=

−

2

x

+

9

⇒

y

=

−

2

3

x

+

3

←

in slope-intercept form

with slope m  

=

−

2

3

Given a line with slope then the slope of a line

perpendicular to it is

∙

x

m

perpendicular

=

−

1

m

⇒

m

perpendicular

=

−

1

−

2

3

=

3

2

⇒

y

=

3

2

x

+

b

←

is the partial equation

to find b substitute  

(

3

,

−

3

)

into the partial equation

−

3

=

9

2

+

b

⇒

b

=

−

6

2

−

9

2

=

−

15

2

⇒

y

=

3

2

x

−

15

2

←

equation of perpendicular lineExplanation:

the equation of a line in  

slope-intercept form

is.

∙

x

y

=

m

x

+

b

where m is the slope and b the y-intercept

rearrange  

2

x

+

3

y

=

9

into this form

⇒

3

y

=

−

2

x

+

9

⇒

y

=

−

2

3

x

+

3

←

in slope-intercept form

with slope m  

=

−

2

3

Given a line with slope then the slope of a line

perpendicular to it is

∙

x

m

perpendicular

=

−

1

m

⇒

m

perpendicular

=

−

1

−

2

3

=

3

2

⇒

y

=

3

2

x

+

b

←

is the partial equation

to find b substitute  

(

3

,

−

3

)

into the partial equation

−

3

=

9

2

+

b

⇒

b

=

−

6

2

−

9

2

=

−

15

2

⇒

y

=

3

2

x

−

15

2

←

equation of perpendicular line

7 0
3 years ago
Read 2 more answers
Write the following number as ratios of integers
s2008m [1.1K]

Our goal here is to somehow "surgically remove" the repeating part of the number, so let's start by putting the original value in a variable and messing around with it a bit.

We'll let x=7.\overline{6}. We want to cut the 0.\overline{6} bit off completely, so let's create the scalpel that'll let us do that. If x=7.\overline{6}, then we can also say that 10x=76.\overline{6}. Maybe I was lying a bit: the x is our real scalpel here, and 10x is where we'll be making the cut. Mathematically, a "cut" is almost always shorthand for subtraction, so let's see what our operation (cutting x off of 10x) leaves us with:

10x-x=76.\overline{6}-7.\overline{6}\\9x=69

The operation was a success! We can now simply divide either side by 9 to find x = 69/9, which, when reduced by dividing the numerator and denominator by 3, gives us x=23/3

7 0
3 years ago
Chart:
True [87]
I would say B because it only makes since, colleges are looking for whats better in students and you will also work in groups also. So i would go with B :)
5 0
3 years ago
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