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antiseptic1488 [7]
3 years ago
15

How do you convert fractions to a specific base with exponents? Is there a specific way or equation for this? For example, i nee

d to solve 1/2^3-x=2^3x-7 but first i have to make the bases the same. I'm not sure how to get 1/2 equal to 2 with exponents
Mathematics
1 answer:
Lilit [14]3 years ago
8 0

Answer:

\frac{1}{2}  =  {2}^{ - 1}

Step-by-step explanation:

remember the property

{a}^{ - n}  =  \frac{1}{ {a}^{n} }

then u get equation with same base

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Simplify in Exponential form <br> First correct answer gets BRAINLIEST
svlad2 [7]

Answer:

Steps to simplifying fractions

Therefore, 9/9 simplified to lowest terms is 1/1.

Reduce 9/8 to lowest terms

9/8 is already in the simplest form. It can be written as 1.125 in decimal form (rounded to 6 decimal places).Step-by-step explanation:

7 0
3 years ago
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Select the statement that correctly compares two numbers. (GIVING BRAINLIEST)
mars1129 [50]

Answer:

4.13=4.130

If there is a 0 at the end of a decimal then you can drop it.

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Cdef maps to jklm with the transformation (x,y) to (4x,4y) to (x-4,y-9). What is the value of ef/lm.
Elina [12.6K]

Answer:

EF / LM = 1 / 4

Step-by-step explanation:

Transformation is the movement of a point from its initial location to a new location. Types of transformation are reflection, translation, dilation and rotation.

Dilation is the enlargement or reduction in the size of a figure. If a point A(x, y) is dilated by a scale factor of k, the new point is at A'(kx, ky).

Translation is the movement of a point right, left, up or down. If a point A(x, y) is translated a units left and b units down, the new point is at A'(x - a, y - b).

Translation preserves the size and shape of an object. Dilation preserves the shape but not the size.

Cdef maps to jklm with the transformation (x,y) to (4x,4y) to (x-4,y-9).

CDEF was first dilated by a scale factor of 4 to get (4x,4y) before it was translated by (x-4,y-9). Since dilation changes the size of the figure, hence JKLM would be 4 times the size of CDEF. Therefore:

LM / EF = 4

EF / LM = 1 / 4

6 0
3 years ago
(2a+b)/(2a-b) <br><br> / - division sign
MA_775_DIABLO [31]

Answer:

WHAT ON EARTH

Step-by-step explanation:

5 0
3 years ago
50 POINTS FIRST GOOD ANSWER GETS BRAINLIEST
Inessa [10]

Answer:

f(x)=3+|2x-5|=\left\{        \begin{array}{ll}            2x-2& \quad x \geq 5/2 \\            -2x+8 & \quad x < 5/2         \end{array}    \right.

Step-by-step explanation:

We are given the function:

f(x)=3+|2x-5|

Remember that by the definition of absolute value:

\displaystyle |x|= \left\{        \begin{array}{ll}            x & \quad x \geq 0 \\        -    x & \quad x < 0        \end{array}    \right.

Our absolute value is:

|2x-5|

First, we will find when it becomes 0. Set the equation equal to 0:

2x-5=0

Solve for <em>x: </em>

<em />x=5/2<em />

<em />

So, we can see that for all values greater than <em>x </em>= 5/2, 2x - 5 is positive.

For all values less than <em>x </em>= 5/2, 2x - 5 is negative.

Therefore (the positive case go above, and the negative case go below):

|2x-5|= \left\{        \begin{array}{ll}            2x-5 & \quad x \geq 5/2 \\           -(2x-5) & \quad x < 5/2         \end{array}    \right.

Finally, we can add a three:

f(x)=3+|2x-5|=\left\{        \begin{array}{ll}            3+(2x-5) & \quad x \geq 5/2 \\            3+(-(2x-5)) & \quad x < 5/2        \end{array}    \right.

Simplify if desired:

f(x)=3+|2x-5|=\left\{        \begin{array}{ll}            2x-2& \quad x \geq 5/2 \\            -2x+8 & \quad x < 5/2         \end{array}    \right.

5 0
3 years ago
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