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Nookie1986 [14]
2 years ago
10

[(2/3)^2]^3 X (2/3)^2 ÷ (2/3)^8 ​

Mathematics
1 answer:
skad [1K]2 years ago
8 0

Answer:

Reduce the expression, if possible, by cancelling the common factors.[(2/3)^2]^3 X (2/3)^2 ÷ (2/3)^8 ​ = 1

Step-by-step explanation:

4/9^3 x (2/3)^2 ÷ (2/3)^8

4/9^3 x 4/9 ÷ (2/3)^8

4/9^3 x 4/9 divided by 256/6561

64/729 x 4/9 divided by 256/6561

= 1

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The student council has $289 to spend on decorations for the fall festival. If decorations for each table cost $13, how many tab
Luba_88 [7]

Answer:

22.2

Step-by-step explanation:

7 0
4 years ago
100 points pls help me
sashaice [31]

Answer:

156 barrels, 26 minutes

Step-by-step explanation:

The amount of minutes since the worker has started to work: x

Convertor: 26+5x

Worker: 6x

Since the question stated that the number of completed barrels are the same: 26+5x=6x

x=26

It has been 26 minutes and each line has 156 barrels filled out.

Hope this helps, please give brainlest if you are statisfied with this responce-it will help me a lot! Thanks!

6 0
3 years ago
Read 2 more answers
: Below are two expressions. Simplify each and then choose the statement that is true.
fgiga [73]
First, what we should do is simplify both expressions.
 We have then:

 Expression 1:
 
(3x2) 3x ^ 2
 (9x ^ (2 + 2))
 9x ^ 4
 Expression 2:
 (3x ^ 3) ^ 2 (x ^ 2)
 (9x ^ 6) (x ^ 2)
 (9x ^ (6 + 2))
 9x ^ 8
 Answer: 
 The exponents on Expression # 2 are greater than the exponents of Expression # 1.
 (8> 4)
5 0
4 years ago
25 POINTS PLEASE HELP
Digiron [165]

Answer:

all i know is that the perimeter is 24

Step-by-step explanation:

i also believe that the Coordinate on the top left is (-4,4) and the one below is (0,4)

6 0
3 years ago
Read 2 more answers
Use the variation of parameters method to solve the DR y" + y' - 2y = 1
postnew [5]

Answer:

y(t)\ =\ C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}

Step-by-step explanation:

As given in question, we have to find the solution of differential equation

y"+y'-2y=1

by using the variation in parameter method.

From the above equation, the characteristics equation can be given by

D^2+D-2\ =\ 0

=>D=\ \dfrac{-1+\sqrt{1^2+4\times 2\times 1}}{2\times 1}\ or\ \dfrac{-1-\sqrt{1^2+4\times 2\times 1}}{2\times 1}

=>\ D=\ -2\ or\ 1

Since, the roots of characteristics equation are real and distinct, so the complementary function of the differential equation can be by

y_c(t)\ =\ C_1e^{-2t}+C_2e^t

Let's assume that

     y_1(t)=e^{-2t}          y_2(t)=e^t

=>\ y'_1(t)=-2e^{-2t}        y'_2(t)=e^t

   and g(t)=1

Now, the Wronskian can be given by

W=y_1(t).y'_2(t)-y'_1(t).y_2(t)

   =e^{-2t}.e^t-e^t(-e^{-2t})

   =e^{-t}+2e^{-t}

   =3e^{-t}

Now, the particular solution can be given by

y_p(t)\ =\ -y_1(t)\int{\dfrac{y_2(t).g(t)}{W}dt}+y_2(t)\int{\dfrac{y_1(t).g(t)}{W}dt}

=\ -e^{-2t}\int{\dfrac{e^t.1}{3.e^{-t}}dt}+e^{t}\int{\dfrac{e^{-2t}.1}{3.e^{-t}}dt}

=\ -e^{-2t}\int{\dfrac{1}{3}dt}+\dfrac{e^t}{3}\int{e^{-t}dt}

=\dfrac{-e^{-2t}}{3}.t-\dfrac{1}{3}

=-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}

Now, the complete solution of the given differential equation can be given by

y(t)\ =\ y_c(t)+y_p(t)

      =C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}

5 0
3 years ago
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