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katrin [286]
3 years ago
5

What is the equation of the line (in slope-intercept form) that passes through the point (5,−1) and is parallel to the line y=2x

−7
Mathematics
1 answer:
tatyana61 [14]3 years ago
5 0

Answer:

<h2>y = 2x - 11</h2>

Step-by-step explanation:

\text{Let}\\\\k:y=m_1x+b_1\\\\l:y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\\\===============================

\text{The slope-intercept form of an equation of a line:}\\\\y=mx+b\\\\m-slope\\b-y-intercept\\\\==========================

\text{We have the equation of a line:}\ y=2x-7\to m_1=2.\\\\\text{The slope of a parallel line:}\ m_2=m_1=2.\\\\\text{We have the equation:}\\\\y=2x+b\\\\\text{Put the coordinates of the point (5, -1) o the equation:}\\\\-1=2(5)+b\\-1=10+b\qquad\tex\text{subtract 10 from both sides}\\-11=b\to b=-11\\\\\text{Finally:}\\\\y=2x-11

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Answer:

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Given:

Desmond deposits $ 50 monthly.

Yearly he deposits = $50×12 = $ 600

Rate of interest compounded monthly = 4.7%

To find the amount he will receive after 10 years and the rate of change the value of his account after 10 years.

Formula

A = P(1+\frac{r}{(n)(100)} )^{nt}

where,

A be the final amount

P be the principal

r be the rate of interest

t be the time and

n be the number of times the interest is compounded.

Now,

Taking,

P = 600, r = 4.7, n = 12, t = 10 we get,

A = 600(1+\frac{4.7}{(12)(100)}) ^{(12)(10)}

or, A = 600(1.0039)^{120}

or, A = 959.1

Now,

At starting he has $ 600

At the end of 10 years he will be having $ 959.1

So,

The amount of change in his account = $ (959.1-600) = $ 359.1

Therefore the rate of change = (\frac{359.1}{600} )(100) %

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b) The rate of change in his account is 59.85% after 10 years.

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Answer:

E. 22

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We know that b>6, and we are asked to find the possible values for 3b + 3.

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