All categories

3 years ago

Properties of exponents. the answer is 1/2^12 i need help with the work

Mathematics

2 answers:

Vikentia [17]3 years ago

7 0

Step-by-step explanation:

Answer in attached image...

hope it helps

Lady_Fox [76]3 years ago

5 0

(2^-1 -2^-1 × 2^-4)^2

(1/2 - 1/2 × 1/2^4)^2

(1/2 - 1/2^5)^2

(1/2 - 1/32)^2

(16/32-1/32)^2

(15/32)^2

(15/2^5)^2

Must click thanks and mark brainliest

You might be interested in

Scott bought a backpack for $35.99 at Walmart. He needed to pay a 10% tax. How much did the backpack cost in all?

stellarik [79]

$35.99 x .10=3.599
$35.99 + 3.60=$39.59

8 0

4 years ago

Which equation represents a line perpendicular to the line shown on the graph?

Vadim26 [7]

Answer:

See description below.

Step-by-step explanation:

To choose the correct equation, find the slope of the line on the graph. Identify two points on the line. Then subtract to find their rate of change using the slope formula.

$m=\frac{y_2-y_1}{x_2-x_1}$

When you know the slope, find the negative reciprocal. For example, if the slope is 2/1 then the negative reciprocal is -1/2. This is the slope of a perpendicular line to a line with slope 2. Choose the equation which has this same slope.

Example:

y = 2x -1

y= -1/2 +5

4 0

3 years ago

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$