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3 years ago

14

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

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