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2 years ago

[(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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3 years ago

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den301095 [7]

Answer:

Step-by-step explanation:

The best way to do this is to first remove the parenthesis by distributing the negative into them. Then we will group the like terms together, moving their respective signs with them, then we will simplify. For the first one, after distributing the negative into the parenthesis (remember that a negative in front of a set of parenthesis containing a polynomial changes the signs in front of each term within the polynomial; negatives become positives and positives become negatives):

$3x^2-6x+11-10x^2+4x-6$ and then grouping like terms next to each other:

$3x^2-10x^2-6x+4x+11-6$ (this step is not completely necessary; I just encourage new learners to do it so as to not miss any of the like terms)

The above then simplifies to

$-7x^2-2x+5$

The next one, after distributing the negative into the parenthesis:

$-3x^2-5x-3+10x^2+7x-2$ and then grouping like terms next to each other (remember to move their signs with them!):

$-3x^2+10x^2-5x+7x-3-2$ which simplifies to

$7x^2+2x-5$

Now for the last one:

$12x^2+6x-5-5x^2-8x+12$ and

$12x^2-5x^2+6x-8x-5+12$ which simplifies to

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3 years ago

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WITCHER [35]

Answer:

x=28 cm

Step-by-step explanation:

Ratio of length of bigger sides to smaller sides=20:10=2:1.

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4 0

2 years ago

The populations of termites and spiders in a certain house are growing exponentially. The house contains 120 termites the day yo

zysi [14]

Answer:

in order to triple the inicial population of spiders, will take 50395 days

Step-by-step explanation:

we can define the termite population function as T(t) and the one for spiders as S(t) , where t represents time measured in days

since both have and exponencial growth

T(t)= a*e^(b*t)

S(t)= c*e^(d*t)

1) when the day the person moves in , t=0 and T(0)= 120 termites

T(0) = a*e^(b*0) = a = 120

2) after 4 days , t=4 and the house contains T(4) = 210 termites

T(4)= 120*e^(b*4) = 210 → 4*b = ln (210/120) → b = (1/4)* ln(210/120)= 0.14

therefore

T(t) = 120*e^(0.14*t)

3) 3 days after moving in , t=3, there were T(3) = 120*e^(0.14*3)=182.63≈ 182 termites . The number of spiders is half of the number of termites → S(3) = T(3) * 1 spider/ 2 termites =91.31 spiders ≈ 91 spiders

4) after 8 days of moving in , t=8, there were T(8) = 120*e^(0.14*8)=367.78≈ 368 termites . The number of spiders is 0.25 times the number of termites → S(8) = T(8) * 1 spider/ 4 termites =91.94 spiders ≈ 92 spiders

from

S(t)= c*e^(d*t) → d = ln [S (tb)/S (ta) ] / (tb-ta)

therefore d = ln [ S(8)/S(3) ] / (8 - 3 ) = 2.18*10^-5

in order to triple the initial population

S(t3) = 3 *S(0) = 3*[c*e^(d*0)] = 3*c

S(t3) = c*e^(d*t3) = 3* c → t3 = ln(3) / d = 50395 days

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3 years ago

How many solutions does y=(2-7)^2

vladimir1956 [14]

I believe the equation y=(2-7)^2 has 1 solution which is y=25

3 0

3 years ago

[(2/3)^2]^3 X (2/3)^2 ÷ (2/3)^8 ​

[(2/3)^2]^3 X (2/3)^2 ÷ (2/3)^8