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2 years ago

Evaluate 1/6x + 5/18 when x = 1/3

Mathematics

1 answer:

lara31 [8.8K]2 years ago

6 0

Answer:

1/3 is the simplified answer

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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

2 years ago

Please help me on these

Sergeu [11.5K]

1. x = 25 inches
2. Triangle c

3 0

2 years ago

800 students are going on a school trip by bus.

Tema [17]

The answer is twenty four

6 0

3 years ago

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The graph of the equation y=12x−3 Determine the value of y for the point (4, y).

Novosadov [1.4K]

Answer:

$y=45$

Step-by-step explanation:

$y=12x-3$

$y=12(4)-3$

$y=45$

4 0

2 years ago

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Find the area of a parallelogram if a base and corresponding altitude have the indicated lengths.

Vitek1552 [10]

The answer is 3/4 ft² = 0.75 ft²

The area (A) of the parallelogram with altitude a and base b is:
A = a * b

We have:
A = ?
a = 6 in
Since 1 inch is 0.083 feett, then 6 inches is
a = 6 * 0.083 ft = 0.498 ft ≈ 0.5 ft = 1/2 ft
b = 1 1/2 ft = 1 + 1/2 ft = 2/2 + 1/2 ft = 3/2 ft
_______
A = a * b
a = 1/2 ft
b = 3/2 ft
A = 1/2 * 3/2 = 3/4 ft² = 0.75 ft²

4 0

3 years ago