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3 years ago

Write the equation of the line that is parallel to the given segment and that passes through the point (-3,0).

Mathematics

1 answer:

grandymaker [24]3 years ago

5 0

Answer:

Option (B)

Step-by-step explanation:

In the figure attached,

A straight line is passing through two points (0, 2) and (3, 1).

Slope of this line ( $m_1$ ) = $\frac{y_2-y_1}{x_2-x_1}$

= $\frac{2-1}{0-3}$

= $-\frac{1}{3}$

Let the slope of a parallel to the line given in the graph = $m_2$

By the property of parallel lines,

$m_1=m_2$

$m_2=-\frac{1}{3}$

Equation of a line passing through a point (x', y') and slope 'm' is,

y - y' = m(x - x')

Therefore, equation of the parallel line which passes through (-3, 0) and having slope = $-\frac{1}{3}$ will be,

$y-0=-\frac{1}{3}(x+3)$

$y=-\frac{1}{3}x-1$

Option (B). will be the answer.

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Read 2 more answers

Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.

finlep [7]

Answer:

a) $v = \frac{[L]}{[T]} = LT^{-1}$

b) $a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

c) $\int v dt = s(t) = [L]=L$

d) $\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

e) $\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

$v = \frac{ds}{dt}$

$a= \frac{dv}{dt}$

Part a

If we do the dimensional analysis for v we got:

$v = \frac{[L]}{[T]} = LT^{-1}$

Part b

For the acceleration we can use the result obtained from part a and we got:

$a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

Part c

From definition if we do the integral of the velocity respect to t we got the position:

$\int v dt = s(t)$

And the dimensional analysis for the position is:

$\int v dt = s(t) = [L]=L$

Part d

The integral for the acceleration respect to the time is the velocity:

$\int a dt = v(t)$

And the dimensional analysis for the position is:

$\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

$\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

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3 years ago