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3 years ago

Simplify the following rational expression. Use absolute value symbols when needed. 3 √216/1728

Mathematics

1 answer:

bagirrra123 [75]3 years ago

3 0

Answer:

1/2

Step-by-step explanation:

Absolute value symbols are only needed in certain cases when finding the nth root of a number when n is even. For example, the 4th root of (-2) ^ 20 is |(-2) ^ 5| or 32. To find a cube root, work backwards. Find a number whose cube is the radicand.

Notice that the radical expression is of the form n√b, where b is a real number and n is a positive integer. This expression can be written as an exponential equation of the form a^n = b, where a is also a real number.

Write 3 √216/1728 as an exponential equation of the form a^n = b

a^3 = 216/1728

Because n is odd, there is only one value of a that makes the equation true. In this case, a will be a fraction. Determine the numerator of a. Find the number that, when cubed, is 216.

(6)^3 = 216

Now determine the denominator of a. Find the number that, when cubed, is 1728.

(12)^3 = 1728

Use the numerator and denominator found above to complete the equation a^3 = 216/1728. . Since n is odd, absolute value symbols are not necessary.

(6/12)^3 = 216/1728

Finally, simplify this value of a.

6/12 = 1/2

Therefore, 3 √216/1728 = 1/2

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4 years ago

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3 years ago

Find the x-and y-intercepts of the graph of 4x + y = 28. State your answers as

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3 years ago

If y varies jointly with x and y=5 when x=15 and z=24 find y when x=6 and z=14

Scorpion4ik [409]

Answer:

if they vary together then y would be 2

Step-by-step explanation:

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4 years ago

Find the area that the curve encloses and then sketch it.<br> r = 3 + 8 sin(6)

Rudiy27

Answer:

$A=41\pi\: \text{units}^2\approxA\approx128.8053\:\text{units}^2$

Step-by-step explanation:

I assume you mean $r=3+8\sin\theta$ :

Use the formula $\displaystyle A=\int\limits^a_b \frac{1}{2} {r(\theta)^2} \, d\theta$ where $a$ and $b$ are the lower and upper bounds and $r(\theta)$ is the equation of the polar curve.

Since the graph is symmetrical about the line $\displaystyle \theta=\frac{\pi}{2}$ , let the bounds of integration be $\displaystyle \biggr(-\frac{\pi}{2},\frac{\pi}{2}\biggr)$ to find half the area of the curve, and then find twice of that area:

$\displaystyle A=\int\limits^a_b \frac{1}{2} {r(\theta)^2} \, d\theta\\\\A=2\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \frac{1}{2} {(3+8\sin\theta)^2} \, d\theta\\\\A=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} 9+48\sin\theta+64\sin^2\theta \, d\theta\\\\A=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} 9+48\sin\theta+64\biggr(\frac{1-\cos2\theta}{2} \biggr) \, d\theta\\\\\\A=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} (9+48\sin\theta+32-32\cos2\theta) \, d\theta$

$\displaystyle A=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} (41+48\sin\theta-32\cos2\theta) \, d\theta\\\\A=41\theta-48\cos\theta-16\sin2\theta\biggr|^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\\\\$

$A=\biggr[41\biggr(\frac{\pi}{2}\biggr)-48\cos\biggr(\frac{\pi}{2}\biggr)-16\sin2\biggr(\frac{\pi}{2}\biggr)\biggr]-\biggr[41\biggr(-\frac{\pi}{2}\biggr)-48\cos\biggr(-\frac{\pi}{2}\biggr)-16\sin2\biggr(-\frac{\pi}{2}\biggr)\biggr]\\\\A=\biggr[\frac{41\pi}{2}-24\sqrt{2}\biggr]-\biggr[-\frac{41\pi}{2}+24\sqrt{2}\biggr]\\ \\A=41\pi\\\\A\approx128.8053$

Thus, the area of the curve is 41π square units. See below for a graph of the curve and its shaded area.

7 0

3 years ago