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aniked [119]
2 years ago
10

Jethro used dimensional analysis to convert one rate of speed to another. Two of the conversion factors he used are 5280 ft/1 mi

and 1 hr/60 min. What could be the units of the initial rate of speed and the final rate of speed? Justify your answer.
Mathematics
1 answer:
NNADVOKAT [17]2 years ago
3 0

The units of the initial and final rates of speed are;

<u><em>Initial rate of Speed = ft/hr</em></u>

<u><em>Final rate of speed = mi/min</em></u>

  • We are told that the conversion factors he used are;

5280 ft/1 mi and 1 hr/60 min

  • Now, the meaning of 5280 ft/1 mi denotes that;

5280 ft mile is equivalent to 1 mile

This means that the initial distance <em>unit </em>used was ft while the final one was mile

Also, 1 hr/60 min denotes that;

1 hour is equivalent to 60 minutes.

This means that the initial time units used was hour while the final one was minute.

  • In conclusion, since we know the initial time and distance units used, we can say that the initial rate of speed ft/hr while final speed rate was miles/minutes.

Read more at; brainly.com/question/17734333

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<span>5x + y = -21:
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You are given the parametric equations x=2cos(θ),y=sin(2θ). (a) List all of the points (x,y) where the tangent line is horizonta
vladimir1956 [14]

Answer:

The solutions listed from the smallest to the greatest are:

x:  -\sqrt{2}   -\sqrt{2}  \sqrt{2}  \sqrt{2}

y:      -1         1     -1     1

Step-by-step explanation:

The slope of the tangent line at a point of the curve is:

m = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }

m = -\frac{\cos 2\theta}{\sin \theta}

The tangent line is horizontal when m = 0. Then:

\cos 2\theta = 0

2\theta = \cos^{-1}0

\theta = \frac{1}{2}\cdot \cos^{-1} 0

\theta = \frac{1}{2}\cdot \left(\frac{\pi}{2}+i\cdot \pi \right), for all i \in \mathbb{N}_{O}

\theta = \frac{\pi}{4} + i\cdot \frac{\pi}{2}, for all i \in \mathbb{N}_{O}

The first four solutions are:

x:   \sqrt{2}   -\sqrt{2}  -\sqrt{2}  \sqrt{2}

y:     1        -1        1     -1

The solutions listed from the smallest to the greatest are:

x:  -\sqrt{2}   -\sqrt{2}  \sqrt{2}  \sqrt{2}

y:      -1         1     -1     1

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oksano4ka [1.4K]

Answer: the answer is negative one over four

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3 years ago
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GenaCL600 [577]

Answer:

The correct option is:

h = 1, k = 16

Step-by-step explanation:

y=4x^2-8x+20 =0

It is a quadratic formula in standard form:

ax^2+bx+c

where a = 4 , b = -8 and c=20

The vertex form is:

a(x − h)2 + k = 0

h is the axis of symmetry and (h,k) is the vertex.

Calculate h according to the following formula:

h = -b/2a

h= -(-8)/2(4)

h = 8/8

h = 1

Substitute k for y and insert the value of h for x in the standard form:

ax^2+bx+c

k = 4(1)^2+(-8)(1)+20

k = 4-8+20

k=-4+20

k = 16

Thus the correct option is h=1, k=16....

8 0
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Substitute the variable for the value given.

11 - 3m becomes 11 - 3(2)

Now do the order of operations.

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So, the answer is 5.
6 0
2 years ago
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