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12345 [234]
3 years ago
14

Find the slope of the line that contains the points (-4, -1) and (-4,-2).

Mathematics
1 answer:
lys-0071 [83]3 years ago
6 0

Answer: m=0

Step-by-step explanation:

You might be interested in
In converting 10 inches to centimeters, what unit (omit the number) would you place in the numerator of your ratio?
belka [17]

Answer:

  centimeters

Step-by-step explanation:

The numerator of the conversion factor always has the "to" units. The denominator has the "from" units. That way, when you multiply, the "from" units cancel:

  (xx <em>from</em>) · (yy <em>to</em>)/(zz <em>from</em>) = xx·yy/zz · (<em>from/from</em>) · <em>to</em> = xx·yy/zz · <em>to</em>

Here, you want to convert to centimeters, so centimeters will be the units in the numerator.

  10 in · (2.54 cm)/(1 in) = 25.4 cm

_____

The conversion factor is always "1". That is, the numerator and denominator are always <u>equal</u> in value. Here, 2.54 cm = 1 in, so (2.54 cm)/(1 in) = 1. You can multiply by 1 anytime you like. For units conversion, it only has the effect of changing the units.

5 0
3 years ago
PLEASE RESPOND!!!
solniwko [45]

Did you ever get the answer

5 0
3 years ago
Please help me out !
VashaNatasha [74]

Answer:

i dont know lol but anyways

Step-by-step explanation:

5 0
3 years ago
(3x+2)+(2x^2+3x-1)<br><br> Please answer ty:)
Paul [167]
Answer = 2x^2+6x+1

3x+2+2x^2+3x-1
3x+1+2x^2+3x
Now combine like terms
3x+1+2x^2+3x
6x+1+2x^2
Rearrange terms
6x+1+2x^2
2x^2+6x+1
4 0
3 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
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