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4 years ago

Please help.........

Mathematics

1 answer:

faust18 [17]4 years ago

5 0

The answer for PA is 60/13. This is found by
The area of the rectangle is 60, therefore, the big triangle areas are 30 each. We know that the hypotenuse is 13, so therefore, we get
1/2(13*PA) = 30. This then simplifies to
13 * PA = 60, which then turns into
PA = 60/13.

Hope this helps and have a nice day:)

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Suppose the equation for line A is given by 2x = 5y – 25. If line A and B are perpendicular

Ludmilka [50]

The required equation of line B is 2y+5x=-2

The equation of a line in point-slope form is expressed as:

y-y0 = m(x-x0)

m is the slope of the line

(x0, y0) is the point on the line

Given the line A expressed as:

2x = 5y – 25

Rewrite in standard form

5y = 2x + 25

y = 2/5 x + 5

The slope will be 2/5

The slope of the perpendicular line will be;

$M = \frac{-1}{(2/5)}\\M = \frac{-5}{2}$

Substitute the point (-6, 14) and slope -5/2 into the equation above to get the required equation of line B

$y-14=-5/2(x+6)\\2(y-14) = -5(x+6)\\2y-28=-5x-30\\2y+5x=-30+28\\2y+5x=-2\\$

Hence the required equation of line B is 2y+5x=-2

Learn more here: brainly.com/question/13140886

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3 years ago

The velocity of an automobile starting from rest is given by the equation below, where v is measured in feet per second and t is

maria [59]

Answer:

a. At t = 5 s

$a(5)=\frac{1785}{\left(6(5)+17\right)^2}=\frac{1785}{2209}\approx0.808 \frac{ft}{s^2}$

b. At t = 10 s

$a(10)=\frac{1785}{\left(6(10)+17\right)^2}=\frac{255}{847}\approx0.301 \frac{ft}{s^2}$

c. At t = 20 s

$a(20)=\frac{1785}{\left(6(20)+17\right)^2}=\frac{1785}{18769}\approx0.095 \frac{ft}{s^2}$

Step-by-step explanation:

We know that the velocity function is given by

$v(t)=\frac{105t}{6t+17}$

Acceleration is the rate of change of velocity so we take the derivative of the velocity function with respect to time.

$a(t)=\frac{dv}{dt}=\frac{d}{dt} (\frac{105t}{6t+17})$

$\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'\\\\105\frac{d}{dt}\left(\frac{t}{6t+17}\right)\\\\\mathrm{Apply\:the\:Quotient\:Rule}:\quad \frac{d}{{dx}}\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \frac{{\frac{d}{{dx}}f\left( x \right)g\left( x \right) - f\left( x \right)\frac{d}{{dx}}g\left( x \right)}}{{g^2 \left( x \right)}}$

$105\cdot \frac{\frac{d}{dt}\left(t\right)\left(6t+17\right)-\frac{d}{dt}\left(6t+17\right)t}{\left(6t+17\right)^2}\\\\105\cdot \frac{1\cdot \left(6t+17\right)-6t}{\left(6t+17\right)^2}\\\\a(t)=\frac{1785}{\left(6t+17\right)^2}$