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3 years ago

11

Which expression is equivalent to (5^√3^2)^1/3 ?

1 answer:

kiruha [24]3 years ago

5 0

Answer:

\frac{-177}{286}

Explanation:

Given,

\left(\frac{3}{11}\times\frac{5}{6}\right)-\left(\frac{9}{12}\times \frac{4}{3}\right)+\left(\frac{5}{13}\times \frac{6}{15}\right)

ft(\frac{5}{22}\right )-\left(\frac{36}{36}\right)+\left(\frac{2}{13}\right)

{5}{22}\right )-1+\left(\frac{2}{13}\right)

{5\times 13-286+2\times 22}{286}

{65-286+44}{286}

{109-286}{286}

{-177}{286}

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Step-by-step explanation:

the degree is 5

As degree means the largest exponent of a term

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4 years ago

Element X decays radioactively with a half life of 11 minutes. If there are 870 grams of Element X, how long, to the nearest ten

serg [7]

Answer:

It would take 27.5 minutes the element to decay to 154 grams.

Step-by-step explanation:

The decay equation:

$\frac {dN}{dt}\propto -N$

$\Rightarrow \R\frac {dN}{dt}=-\lambda N$

$\Rightarrow \frac {dN}N=-\lambda dt$

Integrating both sides

$\Rightarrow \int \frac {dN}N=\int-\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$ln|N_0|=-\lambda .0+c$

$\Rightarrow c=ln|N_0|$

$ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$

Decay equation:

$ln|\frac{N}{N_0}|=-\lambda t$

Given that, the half life of of element X is 11 minutes.

For half life, $N=\frac12 N_0$ , t= 11 min.

$ln|\frac{N}{N_0}|=-\lambda t$

$\Rightarrow ln|\frac{\frac12N_0}{N_0}|=-\lambda . 11$

$\Rightarrow ln|\frac12}|=-\lambda . 11$

$\Rightarrow -\lambda . 11=ln|\frac12}|$

$\Rightarrow \lambda =\frac{ln|\frac12|}{-11}$

$\Rightarrow \lambda =\frac{ln|2|}{11}$ [ $ln|\frac12|=ln|1|-ln|2|=-ln|2|$ , since ln|1|=0]

N=154 grams, $N_0$ = 870 grams, t=?

$ln|\frac{N}{N_0}|=-\lambda t$

$\Rightarrow ln|\frac{154}{870}|=-\frac{ln|2|}{11}.t$

$\Rightarrow t= \frac{ln|\frac{154}{870}|\times 11}{-ln|2|}$

=27.5 minutes

It would take 27.5 minutes the element to decay to 154 grams.

5 0

3 years ago

Tao wants to check that the expression 3(4x – 6) simplifies to 12x – 18. What is the value when 2 is substituted for x into both

Anika [276]

Keywords:

<em>Equation, simplified, variable, compare </em>

For this case, we have an equation of the form $y = f (x)$ , where $f (x) = 3 (4x - 6)$ . The equation was simplified and we want to find the value of $f (x)$ when the variable $x = 2$ and compare the result of both equations. So:

$f (x) = 3 (4x - 6)$ (1)

We apply distributive property to simplify, taking into account that: $a (b + c) = ab + ac$

$f (x) = 12x-18$ (2)

If $x = 2$ , we substitute in the original equation and in the simplified equation:

1) $f (2) = 3 (4 (2) - 6) = 3 (8-6) = 3 * 2 = 6$

2) $f (2) = 12 (2) -18 = 24-18 = 6$

Thus, when $x = 2$ is substituted in both expressions, the result is 6.

Answer:

For $f (x) = 3 (4x - 6)$ and $f (x) = 12x-18$ , substituting $x = 2$ we have the result is 6.

6 0

4 years ago

Read 2 more answers