What is 2 to the power of 4 over 3 equal to? (5 points)

Mathematics

1 answer:

ryzh [129]3 years ago

4 0

Answer:

The answer is the option C

cube root of $16$

Step-by-step explanation:

Remember that

$a^{\frac{x}{y}} =\sqrt[y]{a^{x}}$

in this problem we have

$a^{\frac{x}{y}}=2^{\frac{4}{3}}$

therefore

$2^{\frac{4}{3}} =\sqrt[3]{2^{4}}=\sqrt[3]{16}$

You might be interested in

The following integral requires a preliminary step such as long division or a change of variables before using the method of par

shtirl [24]

Division yields

$\dfrac{x^4+7}{x^3+2x} = x-\dfrac{2x^2-7}{x^3+2x}$

Now for partial fractions: you're looking for constants a, b, and c such that

$\dfrac{2x^2-7}{x(x^2+2)} = \dfrac ax + \dfrac{bx+c}{x^2+2}$

$\implies 2x^2 - 7 = a(x^2+2) + (bx+c)x = (a+b)x^2+cx + 2a$

which gives a + b = 2, c = 0, and 2a = -7, so that a = -7/2 and b = 11/2. Then

$\dfrac{2x^2-7}{x(x^2+2)} = -\dfrac7{2x} + \dfrac{11x}{2(x^2+2)}$

Now, in the integral we get

$\displaystyle\int\frac{x^4+7}{x^3+2x}\,\mathrm dx = \int\left(x+\frac7{2x} - \frac{11x}{2(x^2+2)}\right)\,\mathrm dx$

The first two terms are trivial to integrate. For the third, substitute y = x ² + 2 and dy = 2x dx to get

$\displaystyle \int x\,\mathrm dx + \frac72\int\frac{\mathrm dx}x - \frac{11}4 \int\frac{\mathrm dy}y \\\\ =\displaystyle \frac{x^2}2+\frac72\ln|x|-\frac{11}4\ln|y| + C \\\\ =\displaystyle \boxed{\frac{x^2}2 + \frac72\ln|x| - \frac{11}4 \ln(x^2+2) + C}$