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chubhunter [2.5K]
3 years ago
13

ewrite the rational exponent as a radical by extending the properties of integer exponents. 2 to the 3 over 4 power, all over 2

to the 1 over 2 power
Mathematics
1 answer:
yawa3891 [41]3 years ago
5 0

\bf ~\hspace{7em}\textit{rational exponents} \\\\ a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} ~\hspace{10em} a^{-\frac{ n}{ m}} \implies \cfrac{1}{a^{\frac{ n}{ m}}} \implies \cfrac{1}{\sqrt[ m]{a^ n}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{2^{\frac{3}{4}}}{2^{\frac{1}{2}}}\implies 2^{\frac{3}{4}}\cdot 2^{-\frac{1}{2}}\implies 2^{\frac{3}{4}-\frac{1}{2}}\implies 2^{\frac{3-2}{4}}\implies 2^{\frac{1}{4}}\implies \sqrt[4]{2^1}\implies \sqrt[4]{2}

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A sequence is defined by the formula f(n + 1) = f(n) – 3. If f(4) = 22, what is f(1)?
julsineya [31]

f(n) = f(n+1) +3

f(1) = f(2) +3

f(2) =f(3) +3

f(3) = f(4) +3

So f(1) = f(n) + 3* (n-1)

Put n=4 here so you found

f(1) = f(4) + 3* 3

= 22 +9

=31 ans




                             hope it helps

5 0
3 years ago
Read 2 more answers
Write an equivalent expression by distributing the "-−" sign outside the parentheses:
charle [14.2K]

Answer:

6.6a+6b-7.1

Step-by-step explanation:

-(-6.6a-6b)-7.1\\\\=-(-6.6a)-(-6b)-7.1\\\\=6.6a+6b-7.1

7 0
3 years ago
√2+1<br> b) Rationalize the denominator:<br> √2+1
Misha Larkins [42]

Answer:

there is not a complete question here ...

\sqrt{2} +1 is not in a denominator to rationalize

if it were you would multiply the rational expression (the fraction)

by \frac{\sqrt{2} -1}{\sqrt{2} -1} the result (in the denominator would end up being "2 -1"

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Step-by-step explanation:

4 0
3 years ago
Use the drop-down menus to complete the statements about factoring 14x2 + 6x – 7x – 3 by grouping.
Stella [2.4K]

Answer:1.  GCF of the group ( 6 x - 3 ) is 3.

2.  The common binomial factor is 2 x - 1.

3.  The factored expression is:  ( 2 x - 1 ) ( 7 x + 3 ).

Step-by-step explanation:

14 x² + 6 x - 7 x - 3 =

= ( 14 x² - 7 x ) + ( 6 x - 3 ) =

= 7 x ( 2 x - 1 ) + 3 ( 2 x - 1 ) =

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4 0
3 years ago
Read 2 more answers
for the school play adult tickets cost 4$ and children tickets cost 2$ natalie is working at the ticket counter and just sold 20
lilavasa [31]
Let us formulate the independent equation that represents the problem. We let x be the cost for adult tickets and y be the cost for children tickets. All of the sales should equal to $20. Since each adult costs $4 and each child costs $2, the equation should be

4x + 2y = 20

There are two unknown but only one independent equation. We cannot solve an exact solution for this. One way to solve this is to state all the possibilities. Let's start by assigning values of x. The least value of x possible is 0. This is when no adults but only children bought the tickets.

When x=0,
4(0) + 2y = 20
y = 10

When x=1,
4(1) + 2y = 20
y = 8

When x=2,
4(2) + 2y = 20
y = 6

When x=3,
4(3) + 2y= 20
y = 4

When x = 4,
4(4) + 2y = 20
y = 2

When x = 5,
4(5) + 2y = 20
y = 0

When x = 6,
4(6) + 2y = 20
y = -2

A negative value for y is impossible. Therefore, the list of possible combination ends at x =5. To summarize, the combinations of adults and children tickets sold is tabulated below:

   Number of adult tickets             Number of children tickets
                  0                                                   10
                  1                                                    8
                  2                                                    6
                  3                                                    4
                  4                                                    2
                  5                                                    0




6 0
3 years ago
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