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adelina 88 [10]
3 years ago
14

I need help on question 6. And 7.

Mathematics
1 answer:
Alexxx [7]3 years ago
7 0

Answer:

6-21

Step-by-step explanation:

6-

if you convert 15.9375 to a decimal it is 1.75

now if you convert 1.75 to inches it is 21 inches.  

for 7 it seems that it got cut off but if you send me a pic i dont mind helpin you.

please make me as brainlest

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1. A card is picked at random from a pack of 52 playing cards. The probability that it is a 6 is?
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Answer:

1.) A - 5/52

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Why is it important to calculate the distance between two points on a cartesian coordinate plane?
Gwar [14]

Answer:

Derived from the Pythagorean Theorem, the distance formula is used to find the distance between two points in the plane. The Pythagorean Theorem,

a

2

+

b

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=

c

2

, is based on a right triangle where a and b are the lengths of the legs adjacent to the right angle, and c is the length of the hypotenuse. The relationship of sides

|

x

2

−

x

1

|

and

|

y

2

−

y

1

|

to side d is the same as that of sides a and b to side c. We use the absolute value symbol to indicate that the length is a positive number because the absolute value of any number is positive. (For example,

|

−

3

|

=

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. ) The symbols

|

x

2

−

x

1

|

and

|

y

2

−

y

1

|

indicate that the lengths of the sides of the triangle are positive. To find the length c, take the square root of both sides of the Pythagorean Theorem.

c

2

=

a

2

+

b

2

→

c

=

√

a

2

+

b

2

It follows that the distance formula is given as

d

2

=

(

x

2

−

x

1

)

2

+

(

y

2

−

y

1

)

2

→

d

=

√

(

x

2

−

x

1

)

2

+

(

y

2

−

y

1

)

2

We do not have to use the absolute value symbols in this definition because any number squared is positive.

A GENERAL NOTE: THE DISTANCE FORMULA

Given endpoints

(

x

1

,

y

1

)

and

(

x

2

,

y

2

)

, the distance between two points is given by

d

=

√

(

x

2

−

x

1

)

2

+

(

y

2

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y

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2

Step-by-step explanation:

4 0
3 years ago
I need help for this, thanks
gulaghasi [49]

(a) Yes all six trig functions exist for this point in quadrant III. The only time you'll run into problems is when either x = 0 or y = 0, due to division by zero errors. For instance, if x = 0, then tan(t) = sin(t)/cos(t) will have cos(t) = 0, as x = cos(t). you cannot have zero in the denominator. Since neither coordinate is zero, we don't have such problems.

---------------------------------------------------------------------------------------

(b) The following functions are positive in quadrant III:

tangent, cotangent

The following functions are negative in quadrant III

cosine, sine, secant, cosecant

A short explanation is that x = cos(t) and y = sin(t). The x and y coordinates are negative in quadrant III, so both sine and cosine are negative. Their reciprocal functions secant and cosecant are negative here as well. Combining sine and cosine to get tan = sin/cos, we see that the negatives cancel which is why tangent is positive here. Cotangent is also positive for similar reasons.

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Step-by-step explanation:

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