Algebra 1

Mathematics

2 answers:

bezimeni [28]3 years ago

7 0

None
No answers in this question

mojhsa [17]3 years ago

4 0

Answer:

None

Step-by-step explanation:

You might be interested in

Find domain of y=rad20-6x

Evgesh-ka [11]

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

5 0

3 years ago

4. Simplify the expression:

Natalija [7]

Answer:

Step-by-step explanation:

$7x^2-6+4x+7-5x^2+x\qquad\text{combine like terms}\\\\=(7x^2-5x^2)+(4x+x)+(-6+7)\\\\=2x^2+5x+1$