1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
DENIUS [597]
2 years ago
14

Simplify each expression. Assume that all variables are positive.

Mathematics
1 answer:
kozerog [31]2 years ago
7 0
Q1. The answer is  \frac{8x^{3}y^{6}  }{27}

( \frac{16 x^{5} y^{10}}{81x y^{2} } )^{ \frac{3}{4} }= ( \frac{16}{81}* \frac{ x^{5} }{x}* \frac{ y^{10} }{y^{2}}   )^{ \frac{3}{4} } \\  \\ 
  \frac{ x^{a} }{ x^{b} }= x^{a-b}  \\  \\ 
( \frac{16}{81}* \frac{ x^{5} }{x}*\frac{ y^{10} }{y^{2}}   )^{ \frac{3}{4} }}=( \frac{16}{81 }* x^{5-1}* y^{10-2})^{ \frac{3}{4} }=( \frac{16}{81 }* x^{4}* y^{8})^{ \frac{3}{4} }= \\  \\ = (\frac{16}{18} )^{ \frac{3}{4} }*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} }=
\frac{(16)^{ \frac{3}{4} }}{(18)^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} }=\frac{( 2^{4} )^{ \frac{3}{4} }}{( 3^{4} )^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} } \\  \\ 
 (x^{a} )^{b} = x^{a*b}  \\  \\ 
\frac{( 2^{4} )^{ \frac{3}{4} }}{( 3^{4} )^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} } =  \frac{ 2^{4* \frac{3}{4} } }{ 3^{4* \frac{3}{4} } } * x^{4* \frac{3}{4} } * y^{8*\frac{3}{4}} = \frac{ 2^{3} }{ 3^{3} } * x^{3} *y^{6} = 
= \frac{8x^{3}y^{6}  }{27}

Q2. The answer is 1/16

(-64) ^ \frac{-2}{3} =(-1* 2^{6} ) ^ \frac{-2}{3}=(-1)^ \frac{-2}{3} *(2^{6} ) ^ \frac{-2}{3} \\\\x^{-a} =  \frac{1}{ x^{a} } \\\\(-1)^ \frac{-2}{3} *(2^{6} ) ^ \frac{-2}{3} = \frac{1}{(-1)^ \frac{2}{3}} *\frac{1}{(2^{6})^ \frac{2}{3}} \\  \\  (x^{a} )^{b}=x^{a*b} \\\\x^{ \frac{a}{b} = \sqrt[b]{ x^{a} } }  \\  \\ 

\frac{1}{(-1)^ \frac{2}{3}} *\frac{1}{2^{6*\frac{2}{3}}} = \frac{1}{ \sqrt[3]{(-1)^{2} } } * \frac{1}{ 2^{4} } =  \frac{1}{ \sqrt[3]{1} } * \frac{1}{16} = \frac{1}{1} * \frac{1}{16}= \frac{1}{16}


Q3. The answer is a^{ \frac{7}{6} }

a^{ \frac{2}{3} } * a^{ \frac{1}{2} }  \\  \\ 
 x^{a}* x^{b}  =x^{a+b}  \\  \\ 
a^{ \frac{2}{3} } * a^{ \frac{1}{2} }= a^{ \frac{2}{3} + \frac{1}{2} } =a^{ \frac{2*2}{3*2} + \frac{1*3}{2*3} }=a^{ \frac{4}{6} + \frac{3}{6} }=a^{ \frac{4+3}{6} }=a^{ \frac{7}{6} }
You might be interested in
I really need help pls!!!
erastova [34]
The answer would be option 1, mass divided by volume
7 0
2 years ago
Read 2 more answers
Two containers are used to hold liquid. These containers have exactly the same shape. The first container has a height of , and
puteri [66]

Step-by-step explanation:

Please specify values so we may answer. Judging by the looks of the question, this uses the concept of similarity

3 0
2 years ago
HELP LAST PROBLEM!
Keith_Richards [23]
Given 4 boxes of pencils cost $5,
Unknown- slope 
$1.25 per box since $5/boxes
Equation y=mx+b
Substitute $1.25/per box for m
Solve y=$1.25x  
The Check
$5=$1.25/per box(4 boxes)
$5=$5
3 0
2 years ago
PLEASE HELP ITS URGENT!!<br> Given: △ABC is isosceles; 1 ≅ 3<br> Prove: AB || CD
Dmitrij [34]

Answer:

AB ║ CD. (Proved)

Step-by-step explanation:

See the attached diagram of the triangle.

It is given that Δ ACD is an isosceles triangle.

Therefore, AC = AD and ∠ ACD = ∠ ADC, ⇒ ∠ 3 = ∠ 4 .......... (1)

Again, given that ∠ 1 = ∠ 3 ........... (2)

Now, from equations (1) and (2) we can write, ∠ 1 = ∠ 4

Now, AB and CD are two straight lines and AD is the transverse line and hence, ∠ 1 and ∠ 4 are alternate angles that are equal.

Therefore, AB and CD are parallel straight lines and AB ║ CD. (Proved)

4 0
3 years ago
For each of the following functions, determine if they are injective. Also determine if theyare surjective. Also determine if th
timurjin [86]

Answer:

Check below

Step-by-step explanation:

1. Definition for intervals

(a,b)=\left \{ x\in\Re : a

2. Functions

1) \Re \rightarrow \Re \\ f(x)=x

Let's perform graph tests.

That's an one to one, injective function. Look how any horizontal line touches that only once. Also, It's a surjective and a bijective one.

2)\Re\geq0\rightarrow\Re , f(x)=x+1\\

Injective, surjective and bijective.

Injective: a horizontal line crosses the graph in one point.

3)f:\Re\geq 0\rightarrow\Re, f(x) = cos(x)

The cosine function is not injective, bijective nor surjective.

4)f:\Re\rightarrow\Re \:f(x)=ex

Since e, is euler number it's a constant. It's also injective, surjective and bijective.

5)  Quite unclear format

6) \:f:\Re\rightarrow(0,\infty), f(x) =ex

Despite the Restriction for the CoDomain, the function remains injective, surjective and therefore bijective.

7) f:\Re\geq 0 \rightarrow  \Re\geq0, f(x) =x^{4}

Not injective nor surjective therefore not bijective too.

8).f:\{-1,2,-3\}}\rightarrow \{1,4,9\}, f(x) =x^{2}

f(-1)=1, f(2)=4, f(-3)=9

Injective (one to one), Surjective,  and Bijective.

10) f:\Re\geq 0\rightarrow [-1,1], f(x)= cos(x)\\-1=cos(x) \therefore x=\pi,3\pi,5\pi,etc.

Surjective.

11.f:R\geq 0[-1,1], f(x) = 0\\

Surjective

12.f: US Citizens→Z, f(x) = the SSN of x.

General function

13.f: US Zip Codes→US States, f(x) = The state that x belongs to.

Surjective

7 0
2 years ago
Other questions:
  • Which operation should be performed last in this problem: 3^2 + 7 × 4? Why?
    13·2 answers
  • 10x -2y = 10 in slope intercept form
    5·1 answer
  • Ms. Fideli can divide all the students in the chorus into equal groups of 12 students each. How many students could be in the ch
    10·2 answers
  • Please help 20 points
    11·1 answer
  • I. Write the answer with the proper label. ( 2 pts each)
    13·1 answer
  • !!!!!!help please!!!!!! Lots of poibts
    11·2 answers
  • What is 0.48 written as a fraction in lowest terms?<br><br><br> I need so much help
    8·1 answer
  • Which expression is equivalent to (3x2+3x−3)−(6x2−4x−2)?
    8·1 answer
  • Use the following information to answer the question that follows:
    11·2 answers
  • Please help <br> Worth 20 points <br> Show the work
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!