Find the equation of a line that is perpendicular to y=−12x−1 and passes through the point (3,2).

Mathematics

1 answer:

Afina-wow [57]3 years ago

4 0

Answer:

y = (1/12)*x + 7/4

Step-by-step explanation:

y=−12x−1

Perpendicular line:

y = a*x + b

y = (1/12)*x + b

2 = (1/12) * 3 + b

2 = (3/12) + b

2 = 1/4 + b

2 - 1/4 = b

7/4 = b

Perpendicular line:

y = (1/12)*x + 7/4

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Emily is observing the velocity of a cyclist at different times. After four hours, the velocity of the cyclist is 14 km/h. After

lina2011 [118]

Assume that Emily starts from rest so that the velocity is zero at 0 hours.

Define the function v(t) for velocity in km/h, and t in hours.

Part A
From t=0 to t=4, v changes from 0 to 14.
Let v(t) = mt + b
The slope is
m = 14/4 = 3.5, and the y-intercept is
b = 0.
Therefore
v(t) = 3.5t, for 0 ≤ t ≤ 4

Fromt=4 to t=9, changes from 14 to 4.
Let v(t) = mt + b
The slope is
m = (4 - 14)/(9 - 4) = -2
Therefore
v = -2t + b
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v(t) = -2t + 22, for 4 < t ≤ 9.

Answer:
The equation for the velocity is
v(t) = 3.5t, 0 ≤ t ≤ 4
= -2t + 22, 4 < t ≤ 9

Part B
The equation for velocity is a piecewise function that is graphed as shown below. If we assume that the second part of the equation is valid at t=10, then
v(10) = -2*10 + 22 = 2 km/h

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Mario has a box of blocks 8 inches long, 6 inches wide, and 7 inches high.

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Answer:

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Find the area of an equilateral triangle with apothem 7 cm.

lbvjy [14]

So hmm check the picture below

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$\bf \textit{area of an equilateral triangle}\\\\
A=\cfrac{s^2\sqrt{3}}{4}\qquad 
\begin{cases}
s=\textit{length of one side}\\
----------\\
s=14\sqrt{3}
\end{cases}\implies A=\cfrac{(14\sqrt{3})^2\sqrt{3}}{4}
\\\\\\
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