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evablogger [386]
3 years ago
10

Simplify √2/3√2 (it's asking for an easier way to solve this) i think it's 2 3/2

Mathematics
2 answers:
ValentinkaMS [17]3 years ago
7 0

Answer:

  2^(1/6)

Step-by-step explanation:

  \dfrac{\sqrt{2}}{\sqrt[3]{2}}=2^{1/2-1/3}=2^{1/6}

Korvikt [17]3 years ago
3 0

Answer:

2 to the power 1/6.

Step-by-step explanation:

√2 / ∛2

= 2^ (1/2) / 2^(1/3)

= 2^(1/2 - 1/3)

= 2^(1/6).

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What is the solution to this equation? <br> log_7 (5 + 11x) = 2
11111nata11111 [884]

Answer:

x = 4

Step-by-step explanation:

Using the rule of logarithms

log_{b} x = n , then x = b^{n}

Given

log_{7} (5 + 11x) = 2 , then

5 + 11x = 7² = 49 ( subtract 5 from both sides )

11x = 44 ( divide both sides by 11 )

x = 4

5 0
3 years ago
What is another way to write the calculation add 16 and 25, and then multiply by 47?
GarryVolchara [31]

Answer:

(16+25)×47

Step-by-step explanation:

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8 0
3 years ago
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Leto [7]
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8 0
2 years ago
The smiths have two children. The sum of their ages is 23. The produce of their ages is 132. How old are the children?
kolezko [41]

For this case we propose a system of equations:

x: Let the variable representing the age of the first child of the Smiths

y: Let the variable representing the age of the second child of the Smiths

According to the data of the statement we have to:

x + y = 23\\x * y = 132

From the first equation we have to:

x = 23-y

We substitute in the second equation:

(23-y) * y = 132\\23y-y ^ 2 = 132\\y ^ 2-23y + 132 = 0

We find the solutions by factoring:

We look for two numbers that, when multiplied, result in 132 and when added, result in 23. These numbers are 11 and 12.

Thus, we have that the factorized equation is:

(y-11) (y-12) = 0

Thus, the solutions are:y_ {1} = 11\\y_ {2} = 12

So, we can take any of the solutions:

With y = 11

Thenx = 23-11 = 12

Therefore, the ages of the children are 11 and 12 respectively.

Answer:

 The ages of the children are 11 and 12 respectively.

6 0
3 years ago
Twice the smallest of three consecutive odd integers is nine more than the largest. find the integers.
polet [3.4K]
 let   x        x +2       x+4     three consecutive odd <span>integers


2x =9+ x+4   </span>Twice the smallest<span> is nine more than the largest 
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x= 13  the first 
the second integer 15   the third  17 </span>
7 0
3 years ago
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