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3 years ago

How do you raise a number to a power other than two

Mathematics

1 answer:

Lynna [10]3 years ago

7 0

Answer:

The most straightforward way to convert a positive power of two into the form 2n is to count the number n of divisions by 2 that it takes to reach a quotient of 1. For example, the number 524,288 requires 19 divisions to reach 1, giving 219: 524,288/2 = 262,144.Nov 14, 2008

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Select the steps for solving the equation need one more already have two

WARRIOR [948]

Answer:

11 - 6x - 12 = -13 -2x

- 6x - 1 = -13 - 2x

-4x - 1 = -13

8 0

4 years ago

Can someone please tell me how to do equal values?

natulia [17]

Answer:

I think you just plug in the first equation to the or the second to the first, it’s actually does not matter. After that you just solve it. For example: we have two equations.

y=11-2x

y=x+2

We just plug in the first equation to the second one.

11-2x=x+2

-3x=9

x=-3. So this is the answer. Hope this helps.

Step-by-step explanation:

3 0

3 years ago

68 pages read in 40 minutes

VikaD [51]

Answer:

1.7 pages per minute

Step-by-step explanation:

divide 68 by 40 and you get 1.7

5 0

3 years ago

1 Resolver X: X-11 = 28 =

cestrela7 [59]

x = 39 is the answer I think

3 0

3 years ago

Please help i will mark brainliest

Evgen [1.6K]

Answer:

2. $\left[\begin{array}{ccc}1&4\\0&3\\\end{array}\right]$

3. $\left[\begin{array}{ccc}-3&21&60\\-15&9&-45\\\end{array}\right]$

4. $\left[\begin{array}{ccc}6&-14\\-2&-6\\\end{array}\right]$

Step-by-step explanation:

2. This matrix is easy, as it just requires addition.

$\left[\begin{array}{ccc}1&4\\0&3\\\end{array}\right]$ + $\left[\begin{array}{ccc}0&0\\0&0\\\end{array}\right]$ = $\left[\begin{array}{ccc}1&4\\0&3\\\end{array}\right]$

3. This matrix requires for the matrices to be multiplied first, then added.

$\left[\begin{array}{ccc}9&-15&18\\-27&15&-9\\\end{array}\right]$ + $\left[\begin{array}{ccc}-12&36&42 \\12& -6 & -36 \\\end{array}\right]$ = $\left[\begin{array}{ccc}-3&21&60\\-15&9&-45\\\end{array}\right]$

4. Here we can add the last 2 matrices to find x.

$\left[\begin{array}{ccc}2&-8\\-4&2\\\end{array}\right]$ + $\left[\begin{array}{ccc}4&-6\\2&-8 \\\end{array}\right]$ = $\left[\begin{array}{ccc}6&-14\\-2&-6\\\end{array}\right]$

Hope this helps! (Please consider giving brainliest)

4 0

3 years ago