All categories

4 years ago

Rewrite the radical expression as an expression with rational exponents.

Mathematics

2 answers:

never [62]4 years ago

8 0

Answer:

D. $5\sqrt{x^7} = x^{\frac{7}{5}$

Step-by-step explanation:

amna04352 shows it above :)

Lostsunrise [7]4 years ago

5 0

Answer:

x to the seven fifths power

Step-by-step explanation:

x to the power 7: x⁷

Fifth root: (x⁷)^⅕ = x^(7/5)

You might be interested in

Jakes family was on vacation for two weeks and five days. Seth’s family was on vacation for three weeks and two days.How many fe

n200080 [17]

$2 \times 7 = 14 \\ + 5 = 19 \\ \\ 3 \times 7 = 21 \\ + 2 = 23 \\ \\ 23 - 19 = 4$

7 0

3 years ago

Please help!! Due really soon!

Alisiya [41]

9514 1404 393

Answer:

f(x) = 6x +1

Step-by-step explanation:

Differences in x-values (first row) are 1, 1, 1.

Differences in y-values (second row) are 6, 6, 6.

The constant ratio of differences (6/1) tells you the function is linear, and has a slope of m = 6/1 = 6.

Using the first point in the form ...

y = mx + b

we have ...

y = 6x + b

7 = 6·1 + b . . . . (x, y) = (1, 7)

1 = b . . . . . . subtract 6

Then the equation can be written ...

y = 6x +1

In functional form, this is ...

f(x) = 6x +1

4 0

3 years ago

The graph shows the cost of buying beets at a farm stand.. (question continued on image)

Vinvika [58]

Answer/Step-by-step explanation:

Part A:

The slope of the line showing proportional relationship can be found using any one point, (2, 5), on the line.

Slope = y/x = ⁵/2

Part B:

Abby is wrong. From the graph, we see that the unit rate = 5/2 = 2.5. This implies that beets cost $2.5 per pound.

4 0

3 years ago

Which of the binomials below is a factor of this trinomial? x^2-16x+28

Leto [7]

(x - 14) (x - 2) is the answer.

3 0

4 years ago

Determine whether the integral is convergent or divergent. ∫[infinity] 2 e^−1/x / x^2 dx : O Convergent O divergent If it is con

monitta

Let $f(x)=e^{-1/x}$ . Then $f'(x)=\frac1{x^2}e^{-1/x}>0$ for all $x\ge2$ , so $f$ is strictly increasing. As $x\to\infty$ , $e^{-1/x}\to e^0=1$ , so $f$ is bounded above by 1. This is to say,

$e^{-1/x}$

and the integral of $\frac1{x^2}$ converges over the same domain, so this integral must also converge by comparison.

We have, by setting $y=-\frac1x$ ,

$\displaystyle\int_2^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx=\int_{-1/2}^0e^y\,\mathrm dy=e^0-e^{-1/2}=1-\frac1{\sqrt e}$

8 0

3 years ago