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Stels [109]
3 years ago
15

Find domain of y=rad20-6x

Mathematics
1 answer:
Evgesh-ka [11]3 years ago
5 0
Domain, on an "even root" context, means, an even root cannot have a negative radicand, since  say for example \bf \sqrt{-25}\ne -5\qquad why?\implies (-5)(-5)=25

so... you end up with an "imaginary value"

so... for the case of 20-6x
"x", the domain, or INPUT
can afford to have any value, so long it doesn't make the radicand negative

to check for that, let us make the expression to 0, and see what is "x" then

\bf 20-6x=0\implies 20=6x\implies \cfrac{20}{6}=x\implies \cfrac{10}{3}=x

now, if "x" is 10/3, let's see \bf \sqrt{20-6\left( \frac{10}{3} \right)}\implies \sqrt{20-20}\implies \sqrt{0}\implies 0

now, 0 is not negative, so the radicand is golden

BUT, if "x" has a value higher than 10/3, the radicand turns negative
for example  \bf x=\frac{11}{3}
\\\\\\

\sqrt{20-6\left( \frac{11}{3} \right)}\implies \sqrt{20-22}\implies \sqrt{-2}

so.. that's not a good value for "x"

thus, the domain, or values "x" can safely take on, are all real numbers from 10/3 onwards, or to infinity if you wish
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What is 15% of 135, show your work
SIZIF [17.4K]

Answer:

20.25

Step-by-step explanation:

Percentage solution with steps:

Step 1: Our output value is 135.

Step 2: We represent the unknown value with $x$

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Step 3: From step 1 above,$135=100\%$

.

Step 4: Similarly, $x=15.\%$

.

Step 5: This results in a pair of simple equations:

$135=100\%(1)$.

$x=15.\%(2)$

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Step 6: By dividing equation 1 by equation 2 and noting that both the RHS (right hand side) of both

equations have the same unit (%); we have

$\frac{135}{x}=\frac{100\%}{15.\%}$

Step 7: Again, the reciprocal of both sides gives

$\frac{x}{135}=\frac{15.}{100}$

$\Rightarrow x=20.25$Therefore, $15.\%$ of $135$ is

sorry if it took to long have a great day and brainliest is appreciated!!!!!

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