Simplify 3.3l : 900ml

1: Create a single variable linear equation that has no solution. Solve the equation algebraically to prove that it does not hav

OLga [1]

Answer:

(1) y=C, $C\neq0$

(2) x=C

(1) y=0

Step-by-step explanation:

For the linear equation, the power on the solitary variable must be 1.

When the graph of a function crosses/ touches the x-axis than the equation has solution/solutions otherwise it does not has a solution.

When it touches/cresses the x-axis for infinite times then it will have an infinite number of solutions.

(1) The linear equation y-C=0, where y is a variable and C is a non-zero constant, $C \neq 0$ , has the graph parallel to the x-axis. It has no solution as the graph will never cross/touch the x-axis.

Algebraically, in linear equation y-C=0, is independent of x, so there is no value of x for the solution to be exist.

(2) The linear equation x-C=0, where x is a variable and C is a constant, has one solution.

The solution for the equation is

x-C=0

$\Rightarrow x=C$ , where $C\neq0$ .

(3) The linear equation y=0, where y is a variable, has the graph coinciding with the x-axis. So, it has infinitely many solutions.

Algebraically, in linear equation y=0, is independent of x, so for all value of x, the given equation is zero. Hence, there are infinitely many solutions.

3 0

3 years ago

How do i simplify it?

Lemur [1.5K]

Answer:

6 3/4x

Step-by-step explanation:

If you follow PEMDAS (Parenthesis, Exponent, Multiply, Divide, Add, Subtract) then you do what is inside the parenthesis first.

3 1/2 divided by 2= 1 3/4

1 3/4 plus 5= 6 3/5

Then, you add the x and since there is no other variable like x, then you just leave it with 6 3/4.

Hope that helped! :)

3 0

2 years ago

Find the directional derivative of f(x,y,z)=2z2x+y3f(x,y,z)=2z2x+y3 at the point (−1,4,3)(−1,4,3) in the direction of the vector

Feliz [49]

$f(x,y,z)=2z^2x+y^3$

$f$ has gradient

$\nabla f(x,y,z)=2z^2\,\vec\imath+3y^2\,\vec\jmath+4xz\,\vec k$

which at the point (-1, 4, 3) has a value of

$\nabla f(-1,4,3)=18\,\vec\imath+48\,\vec\jmath-12\,\vec k$

I'm not sure what the given direction vector is supposed to be, but my best guess is that it's intended to say $\vec u=15\,\vec\imath+25\,\vec\jmath$ , in which case we have

$\|\vec u\|=\sqrt{15^2+25^2}=5\sqrt{34}$

Then the derivative of $f$ at (-1, 4, 3) in the direction of $\vec u$ is

$D_{\vec u}f(-1,4,3)=\nabla f(-1,4,3)\cdot\dfrac{\vec u}{\|\vec u\|}=\boxed{\dfrac{294}{\sqrt{34}}}$

4 0

3 years ago

What is a parallel line to y= -2/3x+3

Natali5045456 [20]

Answer:

-2/3

Step-by-step explanation:

because this is in slope intercept form, m is the slope, so in this case -2/3 is the slope

parallel lines always have the same slope and different y-intercepts

a parallel line could be y=-2/3 +15

4 0

3 years ago

The square rootof which number is rational a7. b 1.96. c. 0.04. d. 13

artcher [175]

Answer:

B and C

Step-by-step explanation:

We need to find which one of these numbers is a perfect square. We know 7 and 13 can be eliminated since they're prime numbers so they can't be perfect squares. B and C are both perfect squares so that's the answer.

3 0

3 years ago

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