All categories

2 years ago

(4x10^2) - (2x10^-1)

Mathematics

1 answer:

balandron [24]2 years ago

7 0

Answer:

Step-by-step explanation:

The LCD here is 10.

Multiplying numerator and denominator of (4*10^2) by 10 results in:

40*10^3

-------------

and rewriting (2x10^-1) with positive exponent in the denominator results in:

---------

So, before simplification, this difference comes out to be:

40000 - 2 39998 19999

----------------- = ---------------- or ------------

10 10 5

You might be interested in

HELP.PLease help.I will if you a briainlest give me the right answer

JulsSmile [24]

A⁣⁣⁣⁣nswer i⁣⁣⁣s i⁣⁣⁣n a p⁣⁣⁣hoto. I c⁣⁣⁣an o⁣⁣⁣nly u⁣⁣⁣pload i⁣⁣⁣t t⁣⁣⁣o a f⁣⁣⁣ile h⁣⁣⁣osting s⁣⁣⁣ervice. l⁣⁣⁣ink b⁣⁣⁣elow!

bit. $^{}$ ly/3a8Nt8n

7 0

2 years ago

Read 2 more answers

23q−34) and (−16q−r)

faltersainse [42]

Answer:

r = 7 q - 34

Step-by-step explanation:

Solve for r:

-34 + 7 q - r = 0

Subtract 7 q - 34 from both sides:

-r = 34 - 7 q

Multiply both sides by -1:

Answer: r = 7 q - 34

7 0

3 years ago

Read 2 more answers

A source of information randomly generates symbols from a four letter alphabet {w, x, y, z }. The probability of each symbol is

koban [17]

The expected length of code for one encoded symbol is

$\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha\ell_\alpha$

where $p_\alpha$ is the probability of picking the letter $\alpha$ , and $\ell_\alpha$ is the length of code needed to encode $\alpha$ . $p_\alpha$ is given to us, and we have

$\begin{cases}\ell_w=1\\\ell_x=2\\\ell_y=\ell_z=3\end{cases}$

so that we expect a contribution of

$\dfrac12+\dfrac24+\dfrac{2\cdot3}8=\dfrac{11}8=1.375$

bits to the code per encoded letter. For a string of length $n$ , we would then expect $E[L]=1.375n$ .

By definition of variance, we have

$\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2$

For a string consisting of one letter, we have

$\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha{\ell_\alpha}^2=\dfrac12+\dfrac{2^2}4+\dfrac{2\cdot3^2}8=\dfrac{15}4$

so that the variance for the length such a string is

$\dfrac{15}4-\left(\dfrac{11}8\right)^2=\dfrac{119}{64}\approx1.859$

"squared" bits per encoded letter. For a string of length $n$ , we would get $\mathrm{Var}[L]=1.859n$ .

5 0

2 years ago

What is an equation of the line that passes through the point (-7,-8) and is

myrzilka [38]

Answer:

all work is shown and pictured

5 0

2 years ago

Read 2 more answers

What single decimal multiplier would you use to increase by 15% followed by a 14% increase?

umka2103 [35]

9514 1404 393

Answer:

1.311

Step-by-step explanation:

That multiplier is the product of the multipliers for each of the increases:

(1 +15%)(1 +14%) = 1.15×1.14 = 1.311

5 0

2 years ago