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Sati [7]
3 years ago
5

Are the following lines parallel, perpendicular, or neither?

Mathematics
1 answer:
natta225 [31]3 years ago
8 0

Answer:

Perpendicular

Step-by-step explanation:

Parallel lines mean lines that have same slope but since both equations have different slopes which you can check by looking at m-value in y = mx + b. In this case m1 or first slope is -4/3 and m2 or second slope is 3/4.

Perpendicular means that both lines are reciprocal to each other. This means the perpendicular condition is or satisfies \displaystyle \large{m_1m_2 = -1 \longrightarrow m_1 = -\frac{1}{m_2} \longrightarrow m_2 = -\frac{1}{m_1}}

We have m1 = -4/3 and m2 = 3/4.

Therefore, -4/3 * 3/4 = -1 thus both lines are perpendicular to each other as it satisfies m1m2 = -1 condition.

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the first simple interest will be $1.73 more than the second one.

<h3>Which is the difference between the two interests?</h3>

The loan is of $575, and there are two options:

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Remember that the simple interest formula is:

SI = (P*R*T)/100

Where:

P = principal value.

R = rate (this is the percentage per year)

T = time, in years.

Then for the first option. the interest will be:

SI = ($575*4.5*7)/100 = $181.13

For the second option we will have the interest:

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The difference is:

$181.13 - $179.4 = $1.73

This means that the first simple interest will be $1.73 more than the second one.

If you want to learn more about simple interests:

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2 years ago
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Answer:

\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}

Step-by-step explanation:

<u>The Derivative of a Function</u>

The derivative of f, also known as the instantaneous rate of change, or the slope of the tangent line to the graph of f, can be computed by the definition formula

\displaystyle f'(x)=\lim\limits_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}

There are tables where the derivative of all known functions are provided for an easy calculation of specific functions.

The derivative of the inverse tangent is given as

\displaystyle (tan^{-1}u)'=\frac{u'}{1+u^2}

Where u is a function of x as provided:

y=3tan^{-1}(x+\sqrt{1+x^2})

If we set

u=(x+\sqrt{1+x^2})

Then

\displaystyle u'=1+\frac{2x}{2\sqrt{1+x^2}}

\displaystyle u'=1+\frac{x}{\sqrt{1+x^2}}

Taking the derivative of y

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Using the change of variables

\displaystyle y'=3[tan^{-1}u]'=3\frac{u'}{1+u^2}

\displaystyle y'=3\frac{u'}{1+u^2}=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+(x+\sqrt{1+x^2})^2}

Operating

\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+x^2+2x\sqrt{1+x^2}+1+x^2}

\boxed{\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}}

8 0
3 years ago
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