I really need help i keep getting confused

(2^8 x 3^−5 x 6^0)−2 x 3 to the power of negative 2 over 2 to the power of 3, whole to the power of 4 x 2^28

Veronika [31]

$(2^8\cdot3^{-5}\cdot6^0)^{-2}\cdot\left(\dfrac{3^{-2}}{2^3}\right)^4\cdot2^{28}\\\\=(2^8)^{-2}\cdot(3^{-5})^{-2}\cdot1^{-2}\cdot\dfrac{(3^{-2})^4}{(2^3)^4}\cdot2^{28}\\\\=2^{-16}\cdot3^{10}\cdot\dfrac{3^{-8}}{2^{12}}\cdot2^{28}\\\\=2^{-16}\cdot2^{28}\cdot2^{-12}\cdot3^{10}\cdot3^{-8}\\\\=2^{-16+28+(-12)}\cdot3^{10+(-8)}\\\\=2^0\cdot3^2=1\cdot9=\boxed{9}\\\\Used:\\\\(a\cdot b)^n=a^n\cdot b^n\\\\(a^n)^m=a^{n\cdot m}\\\\a^n\cdot a^m=a^{n+m}\\\\a^{-n}=\dfrac{1}{a^n}$

4 0

3 years ago

Read 2 more answers

A line segment has endpoints Q(-16, -12) and R(21, -8).

gulaghasi [49]

Answer:

(2.5, - 10 )

Step-by-step explanation:

given endpoints (x₁, y₁ ) and (x₂, y₂ ) then the midpoint is

( $\frac{x_{1}+x_{2} }{2}$ , $\frac{y_{1}+y_{2} }{2}$ )

here (x₁, y₁ ) = (- 16, - 12 ) and (x₂, y₂ ) = (21, - 8 ) , then

midpoint = ( $\frac{-16+21}{2}$ , $\frac{-12-8}{2}$ ) = ( $\frac{5}{2}$ , $\frac{-20}{2}$ ) = (2.5, - 10 )

4 0

2 years ago

What is the solution to this equation? 3x- x+8+5x-2=10

VashaNatasha [74]

3x -x + 8 + 5x - 2 = 10
3x - x + 5x = 10 -8 +2
7x = 4
x = 7/4

5 0

3 years ago

There are 3 islands A,B,C. Island B is east of island A, 8 miles away. Island C is northeast of A, 5 miles away and northwest of

Nostrana [21]

Answer:

The bearing needed to navigate from island B to island C is approximately 38.213º.

Step-by-step explanation:

The geometrical diagram representing the statement is introduced below as attachment, and from Trigonometry we determine that bearing needed to navigate from island B to C by the Cosine Law:

$AC^{2} = AB^{2}+BC^{2}-2\cdot AB\cdot BC\cdot \cos \theta$ (1)