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3 years ago

Order the ratios from least to greatest. 5:3, 6:4, 11/3, 3 to 1, 33:100

Mathematics

1 answer:

guajiro [1.7K]3 years ago

7 0

Answer:

33:100- 6:4- 5:3- 11/3

Step-by-step explanation:

First turn into even fractions

5:3= 1 2/3

6:4=1 1/2

11/3= 3 2/3

3/1= 3

33:100= 1/3

Then look at the numbers.

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An airliner maintaining a constant elevation of 2 miles passes over an airport at noon traveling 500 mi/hr due west. At 1:00 PM,

butalik [34]

Answer:

$\frac{ds}{dt}\approx 743.303\,\frac{mi}{h}$

Step-by-step explanation:

Let suppose that airliners travel at constant speed. The equations for travelled distance of each airplane with respect to origin are respectively:

First airplane

$r_{A} = 500\,\frac{mi}{h}\cdot t\\r_{B} = 550\,\frac{mi}{h}\cdot t$

Where t is the time measured in hours.

Since north and west are perpendicular to each other, the staight distance between airliners can modelled by means of the Pythagorean Theorem:

$s=\sqrt{r_{A}^{2}+r_{B}^{2}}$

Rate of change of such distance can be found by the deriving the expression in terms of time:

$\frac{ds}{dt}=\frac{r_{A}\cdot \frac{dr_{A}}{dt}+r_{B}\cdot \frac{dr_{B}}{dt}}{\sqrt{r_{A}^{2}+r_{B}^{2}} }$

Where $\frac{dr_{A}}{dt} = 500\,\frac{mi}{h}$ and $\frac{dr_{B}}{dt} = 550\,\frac{mi}{h}$ , respectively. Distances of each airliner at 2:30 PM are:

$r_{A}= (500\,\frac{mi}{h})\cdot (1.5\,h)\\r_{A} = 750\,mi$

$r_{B}=(550\,\frac{mi}{h} )\cdot (1.5\,h)\\r_{B} = 825\,mi$

The rate of change is:

$\frac{ds}{dt}=\frac{(750\,mi)\cdot (500\,\frac{mi}{h} )+(825\,mi)\cdot(550\,\frac{mi}{h})}{\sqrt{(750\,mi)^{2}+(825\,mi)^{2}} }$

$\frac{ds}{dt}\approx 743.303\,\frac{mi}{h}$

6 0

3 years ago

The graph of a proportional relationship contains the point (−12,3)(−12,3) . What is the corresponding equation? Enter your answ

DIA [1.3K]

To find your answer you need to do y divided by x.
since x = -12 and y = 3 you do 3 ÷ -12 and you get -0.25, this as a fraction is -1/4 and this is your answer!
Hope this helped!

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3 years ago

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svetlana [45]

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