Ask question

Ask question

All categories

Natali5045456 [20]

3 years ago

14

Identify the pair of exponents that would make the statement FALSE

1 answer:

IgorLugansk [536]3 years ago

6 0

Answer:

The answer of this question is ( D) 2, 10.

You might be interested in

What is the sixth term of (x + y)8?<br> 56x3y5<br><br> 36x3y5<br><br> 36x4y4<br><br> 56x4y4

Brut [27]

It's either the first or second answer

4 0

3 years ago

Find the pattern and use it to list the nth term in the sequence. <br> 1, 1/32,1/243,1/1024,1/3125

Yuki888 [10]

$1;\ \frac{1}{32};\ \frac{1}{243};\ \frac{1}{1024};\ \frac{1}{3125}\\\\1=\frac{1}{1^5}\\\\\frac{1}{32}=\frac{1}{2^5}\\\\\frac{1}{243}=\frac{1}{3^5}\\\\\frac{1}{1024}=\frac{1}{4^5}\\\\\frac{1}{3125}=\frac{1}{5^5}\\\vdots\\\\a_n=\frac{1}{n^5}\ where\ n\in\mathbb{N^+}$

5 0

3 years ago

What is the value of √144

Marysya12 [62]

Answer:

12!

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Write using only positive exponents. 16x^4z^4/36x-2yz^0

kolbaska11 [484]

Answer:

Step-by-step explanation:

16x^4y^-3z^4 / 36x^-2yz^0

16x^4y^-3z^4 = 16x^4z^4/y^3

36x^-2yz^0 = 36x^-2y(1) =36x^-2y = 36y/x^2

16x^4y^-3z^4 / 36x^-2yz^0

= (16x^4z^4/y^3) / (36y/x^2)

= 16x^4z^4/y^3 * x^2/36y

= (4/9)x^6z^4/y^4

or another way

fist multiply it out

f(x) = 4x^(3/5) - x^(8/5)

now differentiate knowing d/dx(x^n) = n x^(n-1)

to get

4*(3/5) x^(-2/5) - 8/5 x^(3/5)

simplify to get

12/5/x^(2/5) - 8/5 x^(3/5)

If this is what your looking for please give me brainiest, i have done this problem in the past so i know how to solve it :)

5 0

2 years ago

Select True or False for each statement.

SSSSS [86.1K]

$\left( \dfrac 1 {64} \right)^{- 5/6} =64^{5/6} = (\sqrt[6]{64})^5 = 2^5 =32$

TRUE

$\sqrt[5]{36^4}=36^{4/5}$

which surely isn't 36. FALSE

$\sqrt{12} - \dfrac 2 5 \sqrt{75} = 2 \sqrt{3} -\dfrac 2 5 (5) \sqrt{3} = 0$

FALSE

For the fourth one we have a

$\sqrt{98b} + \sqrt{2b}$

which isnt

$10\sqrt{b}$

so this is FALSE.

$\dfrac{1}{(\sqrt 5 - \sqrt 6)^2}$

$= \dfrac{1}{(\sqrt 5 - \sqrt 6)^2} \cdot \dfrac{(\sqrt 5 + \sqrt 6)^2}{(\sqrt 5 + \sqrt 6)^2}$

$= \dfrac{(\sqrt 5 + \sqrt 6)^2}{ ( (\sqrt 5 - \sqrt 6)(\sqrt 5 + \sqrt 6))^2}$

$= \dfrac{(\sqrt 5 + \sqrt 6)^2}{( 5-6)^2}$

$=(\sqrt 5 + \sqrt 6)^2$

No fractions in that one so FALSE.

3 0

3 years ago