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3 years ago

After the expression (x16)3/4 is simplified as much as possible, x is raised to what exponent?

Mathematics

1 answer:

telo118 [61]3 years ago

8 0

Answer:

Step-by-step explanation:

(x^16)^3/4

Apply the law of exponents where (a^b)^c = a^(bc).

x^(16*3/4)

Multiply.

x^(48/4)

Divide.

x^(12)

x is raised to the 12th power.

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Find the differentiation y=root x

siniylev [52]

Answer:

$\frac{dy}{dx}$ = $\frac{1}{2\sqrt{x} }$

Step-by-step explanation:

Differentiate using the power rule

$\frac{d}{dx}$ (a $x^{n}$ ) = na $x^{n-1}$

Given

y = $\sqrt{x}$ = $x^{\frac{1}{2} }$ , then

$\frac{dy}{dx}$ = $\frac{1}{2}$ $x^{-\frac{1}{2} }$ = $\frac{1}{2\sqrt{x} }$

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4 years ago

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Determine the vertical asymptotes for the reciprocal of this function f(x)=x^2 + 3x -28

Gennadij [26K]

Answer:

x = -7 and x = 4.

Step-by-step explanation:

f(x) = x^2 + 3x - 28

= (x + 7)(x - 4).

The reciprocal of f(x) can be written as

1 / (x + 7)(x - 4).

When the denominator is zero we have vertical asymptotes

so they are x = -7 and x = 4.

7 0

2 years ago

Write the equation for the following table: X. Y. 0 3 1 7 2 11 3 15

Vanyuwa [196]

Answer:

$y = 4x + 3$

Step-by-step explanation:

The equation of the table of values given can be written as $y = 4x + 3$ .

When x = 0,

$y = 4(0) + 3$

$y = 0 + 3 = 3$ this corresponds with what we have in the table above.

Let's try another.

When x = 2,

$y = 4(2) + 3$

$y = 8 + 3$

$y = 11$

This also corresponds with what we have in the table.

Therefore, the equation of the table given is $y = 4x + 3$ .

4 0

3 years ago

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Given that y = (e^x)/x, x > 0, find the range of values of x where y is a decreasing

vodka [1.7K]

Answer:

The interval for which y is a decreasing function of x is:

$(0, 1)$

Or as an inequality:

$0$

Step-by-step explanation:

We are given the equation:

$\displaystyle y=\frac{e^x}{x}\, ,x>0$

And we want to find the range of values x for which y is a decreasing function of x.

y is decreasing whenever y' is negative. Find y' using the Quotient Rule:

$\displaystyle y'=\frac{(e^x)'(x)-e^x(x)'}{(x)^2}$

Differentiate:

$\displaystyle y'=\frac{xe^x-e^x}{x^2}$

y is decreasing whenever y' is negative. Thus:

$\displaystyle 0>\frac{xe^x-e^x}{x^2}$

Multiply both sides by x². This is always positive so we do not need to change the sign:

$xe^x-e^x$

Factor:

$e^x(x-1)$

eˣ is always positive. So:

$x-1$

Adding one to both sides produces:

$x$

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In interval notation:

$(0, 1)$

Or as an inequality:

$0$

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3 years ago

Plzzzz help!!! Due tomorrow!!!

-Dominant- [34]

Answer:

Quad One: A

Quad Two: C

Quad Three: B

Quad Four: E

Hope This Helps! Have A Nice Night!!

5 0

3 years ago

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