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3 years ago

Solve and type in a different form by using the given theorems of logarithms. log N^3

Mathematics

2 answers:

Kipish [7]3 years ago

8 0

<h3><u>Answer:</u></h3>

Hence, the different form of log N^3 by using the given theorems of logarithms is:

$\log N^3=3\log N$

<h3><u>Step-by-step explanation:</u></h3>

We have to represent the given logarithmic function in some other form using the property of logarithm.

We are given a logarithmic function as:

$\log N^3$

As we know the property of logarithm that:

$\log m^n=n\log m$

Hence, we can represent our given function as:

$\log N^3=3\log N$

Hence, the different form of log N^3 by using the given theorems of logarithms is:

$\log N^3=3\log N$

timofeeve [1]3 years ago

5 0

Log N^3 = 3 log N

using the theorem log a^2 = 2 log a

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Step-by-step explanation:

Given vectors are $\overrightarrow{u}=(3,4)$ and $\overrightarrow{v}=(1,3)$

Now compute the dot product of u and v:

$\overrightarrow{u}.\overrightarrow{v}=(3,4).(1,3)$

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Now find the magnitude of u and v:

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$|\overrightarrow{v}|=\sqrt{1^2+3^2}$

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$|\overrightarrow{v}|=\sqrt{10}$

To find the angle between the given vectors

$\overrightarrow{u}.\overrightarrow{v}=|\overrightarrow{u}|\overrightarrow{v}|cos\theta$

$\theta=cos^{-1}\left[\frac{\overrightarrow{u}.\overrightarrow{v}}{|\overrightarrow{u}|\overrightarrow{v}|}\right]$

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$\theta=cos^{-1}\left[\frac{3}{\sqrt{10}}\right]$

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$\theta=cos^{-1}\left[\frac{3}{\sqrt{10}}\right]$

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Answer:

arc BC = 60°
m∠ADC = 60°
m∠AEB = 105°
m∠ADP = 45°
m∠P = 60°

Step-by-step explanation:

The sum of arcs of a circle is 360°. The given conditions tell us arc BC ≅ arc AB, so the four arcs of the circle have ratios ...

CB : BA : AD : DC = 2 : 2 : 3 : 5

The sum of ratio units is 2+2+3+5 = 12, so each one stands for 360°/12 = 30°. Then the arc lengths are ...

arc BC = arc BA = 60° . . . . 2 ratio units each

arc AD = 90° . . . . . . . . . . . . 3 ratio units

arc DC = 150° . . . . . . . . . . . .5 ratio units

The inscribed angles are half the measure of the intercepted arcs:

∠ADC = (1/2) arc AC = 1/2(120°) = 60°

∠ADP = 1/2 arc AD = 1/2(90°) = 45°

The angles at E are half the sum of the measures of the intercepted arcs.

∠AEB = (arc AB + arc CD)/2 = (60° +150°)/2 = 105°

Angle P is half the difference of the intercepted arcs.

∠P = (arc BD -arc AD)/2 = (210° -90°)/2 = 120°/2 = 60°

In summary, ...

arc BC = 60°

m∠ADC = 60°

m∠AEB = 105°

m∠ADP = 45°

m∠P = 60°

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