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Vitek1552 [10]
3 years ago
5

How would you write 12-^3 using a positive exponent? 12^3 12^0 12^3/1 1/12^3

Mathematics
1 answer:
Brums [2.3K]3 years ago
6 0

Answer:

\frac{1}{ {12}^{3} }

Step-by-step explanation:

{12}^{ - 3}  =  \frac{1}{ {12}^{3} }  \\

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Make a number line and show all values of x such that x<1/3 and x>0
FromTheMoon [43]

Answer:

<-|---------------|----------------|--------------|->

 0                1/3              2/3             1

Step-by-step explanation:

easy peasy

8 0
3 years ago
Given that 3x-y : x+2y = 2 : 5, Determine the ratio x:y
N76 [4]

Answer:

x : y = 9 : 13

Step-by-step explanation:

Given that

3x - y : x + 2y = 2 : 5

Express the ratio in fractional form, that is

\frac{3x-y}{x+2y} = \frac{2}{5} ( cross- multiply )

5(3x - y) = 2(x + 2y) ← distribute parenthesis on both sides )

15x - 5y = 2x + 4y ( subtract 2x from both sides )

13x - 5y = 4y ( add 5y to both sides )

13x = 9y ( divide both sides by 13 )

x = \frac{9}{13} y ( divide both sides by y )

\frac{x}{y} = \frac{9}{13} , that is

x : y = 9 : 13

7 0
3 years ago
Divide 254 by 11.8. round your answer to the nearest hundredth.
Olin [163]
The answer is 21.52 so I hope that helped
8 0
3 years ago
Read 2 more answers
Which expression represents the sentence "five times a number, n"?
vodomira [7]

Answer:

5n

Step-by-step explanation:

:)

3 0
3 years ago
Graph the hyperbola using the transverse axis, vertices, and co-vertices:
Reptile [31]

See attachment for the graph of the hyperbola 12x^2 - 3y^2 - 108 = 0

<h3>How to graph the hyperbola?</h3>

The equation of the hyperbola is given as:

12x^2 - 3y^2 - 108 = 0

Start by calculating the transverse axis

So, we have:

<u>Transverse axis</u>

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

Where

a = 3 and b = 6

The transverse axis is calculated as:

y = ±b/a(x - h) + k

So, we have:

y = ±6/3(x - 0) + 0

Evaluate the difference and sum

y = ±6/3x

Evaluate the quotient

y = ±2x

This means that the transverse axes are y = 2x and y =-2x

<u>The vertices</u>

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

<u>The co-vertices</u>

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

And

a = 3 and b = 6

The co-vertices are

(h - a, k) and (h + a, k)

So, we have:

(0 - 3, 0) and (0 + 3, 0)

Evaluate

(-3,0) and (3, 0)

See attachment for the graph of the hyperbola

Read more about hyperbola at:

brainly.com/question/26250569

#SPJ1

4 0
1 year ago
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