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3 years ago

25 POINTS! Which equation is equivalent to (1/3)=27^x^+^2

Mathematics

2 answers:

GenaCL600 [577]3 years ago

8 0

Answer:

match bases

remeber if a^b=a^c where a=a then b=c

remember

(x^m)^n=x^(mn)

1/(x^m)=x^-m

base of 3

(1/3)^x=(3^-1)^x=3^(-x)

27^(x+2)=(3^3)^(x+2)=3^(3x+6)

3^(-x)=3^(3x+6)

-x=3x+6

0=4x+6

-6=4x

-3/2=x

Genrish500 [490]3 years ago

7 0

Hint: $\bf log_{{ a}}{{ b}}=y \iff {{ a}}^y={{ b}}$

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Leto [7]

Answer:

180 ft^2

Step-by-step explanation:

Kitchen area is 132 ft^2 and laundry room area (8 ft)(6 ft) = 48 ft^2.

Combining these individual areas, we get the total area: 180 ft^2

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3 years ago

A snack-size bag of M&Ms candies is opened. Inside, there are 12 red candies, 12 blue, 7 green, 13 brown, 3 orange, and 10 y

Anestetic [448]

Answer:

$\frac{1}{4180}$

Step-by-step explanation:

GIVEN: A snack-size bag of M&Ms candies is opened. Inside, there are $12$ red candies, $12$ blue, $7$ green, $13$ brown, $3$ orange, and $10$ yellow. Three candies are pulled from the bag in succession, without replacement.

TO FIND: What is the probability that the first two candies drawn are orange and the third is green.

SOLUTION:

Total candies in the bag $=57$

Probability that first ball is orange, $P(A)=\frac{\text{total orange candies in bag}}{\text{total candies in bag}}$

$=\frac{3}{57}=\frac{1}{19}$

Probability that second ball is orange, $P(B)=\frac{\text{total orange candies in bag}}{\text{total candies in bag}}$

$=\frac{2}{56}=\frac{1}{28}$

Probability that third ball is green, $P(C)=\frac{\text{total green candies in bag}}{\text{total candies in bag}}$

$=\frac{7}{55}$

Now, probability that first two balls are orange and third is green is

$=P(A)\times P(B)\times P(C)$

$=\frac{1}{19}\times\frac{1}{28}\times\frac{7}{55}$

$=\frac{1}{19}\times\frac{1}{4}\times\frac{1}{55}$

$=\frac{1}{4180}$

Hence, probability that first two balls are orange and third is green is $\frac{1}{4180}$

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Answer:

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