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harina [27]
3 years ago
8

F(x)= 5x^6 what is the domain

Mathematics
2 answers:
Nookie1986 [14]3 years ago
6 0
The domain is "the values that x can safely take, without throwing off the expression".

in this case, there are no restrictions to what "x" can be, thus "x" can safely take on all real numbers, and that's the domain.
BabaBlast [244]3 years ago
3 0
(- inf, inf)

polynomial 
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For the equation y= x + 5, find the value of y when
Nina [5.8K]

Answer:

a) 3 b) 5 c) 7 d) 9

Step-by-step explanation:

For this, you want to replace x in the equation y=x+5 with each of the values listed.

For the first one, -2, the equation becomes y=-2+5, which is solved to y=3.

For the second one, 0, the equation becomes y=0+5, which is solved to y=5.

For the third one, 2, the equation becomes y=2+5, which is solved to y=7.

For the last one, 4, the equation becomes y=4+5, which is solved to y=9.

**This content involves solving algebraic equations with a known variable, which you may wish to revise. I'm always happy to help!

5 0
3 years ago
3 friends are sharing 1/2 of a blueberry ....how much pie would each friend get
melisa1 [442]

Answer:

each friend would get 1/6 when you divide 1/2 by 3

Step-by-step explanation:

3 0
3 years ago
What is the derivative of x times squaareo rot of x+ 6?
Dafna1 [17]
Hey there, hope I can help!

\mathrm{Apply\:the\:Product\:Rule}: \left(f\cdot g\right)^'=f^'\cdot g+f\cdot g^'
f=x,\:g=\sqrt{x+6} \ \textgreater \  \frac{d}{dx}\left(x\right)\sqrt{x+6}+\frac{d}{dx}\left(\sqrt{x+6}\right)x \ \textgreater \  \frac{d}{dx}\left(x\right) \ \textgreater \  1

\frac{d}{dx}\left(\sqrt{x+6}\right) \ \textgreater \  \mathrm{Apply\:the\:chain\:rule}: \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx} \ \textgreater \  =\sqrt{u},\:\:u=x+6
\frac{d}{du}\left(\sqrt{u}\right)\frac{d}{dx}\left(x+6\right)

\frac{d}{du}\left(\sqrt{u}\right) \ \textgreater \  \mathrm{Apply\:radical\:rule}: \sqrt{a}=a^{\frac{1}{2}} \ \textgreater \  \frac{d}{du}\left(u^{\frac{1}{2}}\right)
\mathrm{Apply\:the\:Power\:Rule}: \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1} \ \textgreater \  \frac{1}{2}u^{\frac{1}{2}-1} \ \textgreater \  Simplify \ \textgreater \  \frac{1}{2\sqrt{u}}

\frac{d}{dx}\left(x+6\right) \ \textgreater \  \mathrm{Apply\:the\:Sum/Difference\:Rule}: \left(f\pm g\right)^'=f^'\pm g^'
\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(6\right)

\frac{d}{dx}\left(x\right) \ \textgreater \  1
\frac{d}{dx}\left(6\right) \ \textgreater \  0

\frac{1}{2\sqrt{u}}\cdot \:1 \ \textgreater \  \mathrm{Substitute\:back}\:u=x+6 \ \textgreater \  \frac{1}{2\sqrt{x+6}}\cdot \:1 \ \textgreater \  Simplify \ \textgreater \  \frac{1}{2\sqrt{x+6}}

1\cdot \sqrt{x+6}+\frac{1}{2\sqrt{x+6}}x \ \textgreater \  Simplify

1\cdot \sqrt{x+6} \ \textgreater \  \sqrt{x+6}
\frac{1}{2\sqrt{x+6}}x \ \textgreater \  \frac{x}{2\sqrt{x+6}}
\sqrt{x+6}+\frac{x}{2\sqrt{x+6}}

\mathrm{Convert\:element\:to\:fraction}: \sqrt{x+6}=\frac{\sqrt{x+6}}{1} \ \textgreater \  \frac{x}{2\sqrt{x+6}}+\frac{\sqrt{x+6}}{1}

Find the LCD
2\sqrt{x+6} \ \textgreater \  \mathrm{Adjust\:Fractions\:based\:on\:the\:LCD} \ \textgreater \  \frac{x}{2\sqrt{x+6}}+\frac{\sqrt{x+6}\cdot \:2\sqrt{x+6}}{2\sqrt{x+6}}

Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions
\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \  \frac{x+2\sqrt{x+6}\sqrt{x+6}}{2\sqrt{x+6}}

x+2\sqrt{x+6}\sqrt{x+6} \ \textgreater \  \mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c}
\sqrt{x+6}\sqrt{x+6}=\:\left(x+6\right)^{\frac{1}{2}+\frac{1}{2}}=\:\left(x+6\right)^1=\:x+6 \ \textgreater \  x+2\left(x+6\right)
\frac{x+2\left(x+6\right)}{2\sqrt{x+6}}

x+2\left(x+6\right) \ \textgreater \  2\left(x+6\right) \ \textgreater \  2\cdot \:x+2\cdot \:6 \ \textgreater \  2x+12 \ \textgreater \  x+2x+12
3x+12

Therefore the derivative of the given equation is
\frac{3x+12}{2\sqrt{x+6}}

Hope this helps!
8 0
3 years ago
How can you use properties to solve equations with variables on both sides
bija089 [108]
You can use properties to solve equations with variables on both side by simplifying get the variable on one side. solve using inverse operations then check to see if it fits in right .
5 0
3 years ago
10 mL equal to c meters
Andreas93 [3]

This problem can be solved using dimensional analysis.


10mL* \frac{1L}{1000mL} = .01L

8 0
3 years ago
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