What is 805 ten equal to

Determine if the equation below is a function with independent variable x. If so, find the domain. If not, find a value of x to

Andre45 [30]

Answer:

No because it contains points (3,4) and (3,-4). You cannot have a x assigned to more than one y-value if you want it to be a function.

Step-by-step explanation:

A function has one output per input.

If we are trying to determine if the given is a function of x, then x is the input.

However I can get two outputs from plugging in x=3.

$3^2+y^2=25$

$9+y^2=25$

Subtract 9 on both sides:

$y^2=25-9$

$y^2=16$

Take the square root of both sides:

$y=\pm \sqrt{16}$

$y=\pm 4$ .

So input x=3 yields y=4 and y=-4.

Since this input has more than one output then the given is not a function of x.

----Also!

If you graph the equation, it is a circle with radius 5 and center (0,0). So I could I plug in any number for x between -5 and 5 excluding -5 and 5 which would yield only one output each. Since plugging in either one gives:

$(\pm 5)^2+y^2=25$

$25+y^2=25$

Subtract 25 on both sides:

$y^2=25-25$

Simplify:

$y^2=0$

There is only one value y such that when you square it gives you 0. That is 0.

x=5 only gives y=0 and x=-5 only gives y=0.

There is no circle, unless it is a circle with radius 0 which means it really wouldn't be a circle, that is a function.

8 0

3 years ago

In order to solve for x in the equation 3x - 18 = 78, What is the FIRST step?

anygoal [31]

Answer:

add 18 to both sides, giving you 3x = 96

6 0

3 years ago

Please see attachment

Dafna11 [192]

Answer:

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

Step-by-step explanation:

Step(i):-

Given function

$f(x) = \frac{-x}{2x^{2} +1}$ ...(i)

Differentiating equation (i) with respective to 'x'

$f^{l} = \frac{2x^{2} +1(-1) - (-x) (4x)}{(2x^{2}+1)^{2} }$ ...(ii)

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} }$

Equating Zero

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$\frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$2 x^{2}-1 = 0$

$2 x^{2} = 1$

$x^{2} = \frac{1}{2}$

$x = \frac{-1}{\sqrt{2} } , x = \frac{1}{\sqrt{2} }$

Step(ii):-

Again Differentiating equation (ii) with respective to 'x'

$f^{ll}(x) = \frac{(2x^{2} +1)^{2} (4x) - 2(2x^{2} +1) (4x)(2x^{2}-1) }{(2x^{2}+1)^{4} }$

put

$x = \frac{1}{\sqrt{2} }$

$f^{ll} (x) > 0$

The absolute minimum value at $x = \frac{1}{\sqrt{2} }$

Step(iii):-

The value of absolute minimum value

$f(x) = \frac{-x}{2x^{2} +1}$

$f(\frac{1}{\sqrt{2} } ) = \frac{-\frac{1}{\sqrt{2} } }{2(\frac{1}{\sqrt{2} } )^{2} +1}$

on calculation we get

The value of absolute minimum value = - 0.3536

Final answer:-

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

3 0

2 years ago

PLZ HELP I WILL GIVE BRAINLYEST AND 100 POINTS IF YOUR CORRECT

Sav [38]

Answer:

3

Step-by-step explanation:

i double checked

4 0

2 years ago

Read 2 more answers

Find the volume of the cube 1/3 km 1/3 km 1/3 km

guapka [62]

Answer:

The volume of cube=1/27 km^3

Step-by-step explanation:

As The volume of cube =the cube(^3) of the sides.

7 0

3 years ago