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expeople1 [14]
4 years ago
10

Which graph does not represent a function of x?

Mathematics
1 answer:
pantera1 [17]4 years ago
7 0

There is no attachment to your question.

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Find the value of X in the triangle shown below
QveST [7]

Answer:

102

Step-by-step explanation

39'+39'=78'

180-78=102

7 0
2 years ago
What is the term to this?
Leni [432]
Proof by contradiction is a form of proof that establishes the truth or the validity of a proposition, by showing that assuming the proposition to be false leads to a contradiction. I googled this so it may not be correct
5 0
3 years ago
A line joins the point
Harrizon [31]

Intersection of the first two lines:

\begin{cases}5x - 2y + 3 = 0\\4x - 3y + 1 = 0\end{cases}

Multiply the first equation by 4 and the second by 5:

\begin{cases}20x - 8y + 12 = 0\\20x - 15y + 5 = 0\end{cases}

Subtract the two equations:

(20x - 8y + 12)-(20x - 15y + 5)=0 \iff 7y+7=0 \iff y=-1

Plug this value for y in one of the equation, for example the first:

5x - 2\cdot (-1) + 3 = 0\iff 5x+5=0 \iff x=-1

So, the first point of intersection is (-1,-1)

We can find the intersection of the other two lines in the same way: we start with

\begin{cases}x=y\\x=3y+4\end{cases}

Use the fact that x and y are the same to rewrite the second equation as

x=3x+4 \iff 2x=-4 \iff x=-2

And since x and y are the same, the second point is (-2, -2)

So, we're looking for a line passing through (-1,-1) and (-2, -2). We may use the formula to find the equation of a line knowing two of its points, but in this case it is very clear that both points have the same coordinates, so the line must be y=x

In the attached figure, line 5x - 2y + 3 = 0 is light green, line 4x - 3y + 1 = 0 is dark green, and their intersection is point A.

Simiarly, line x=y is red, line x = 3y + 4 is orange, and their intersection is B.

As you can see, the line connecting A and B is the red line itself.

5 0
3 years ago
At what point on the paraboloid y = x2 + z2 is the tangent plane parallel to the plane 3x + 2y + 7z = 2? (if an answer does not
Nikitich [7]
If f(x, y, z) = c represent a family of surfaces for different values of the constant c. The gradient of the function f defined as \nabla f is a vector normal to the surface f(x, y, z) = c.

Given <span>the paraboloid

y = x^2 + z^2.

We can rewrite it as a scalar value function f as follows:

f(x,y,z)=x^2-y+z^2=0

The normal to the </span><span>paraboloid at any point is given by:

\nabla f= i\frac{\partial}{\partial x}(x^2-y+z^2) - j\frac{\partial}{\partial y}(x^2-y+z^2) + k\frac{\partial}{\partial z}(x^2-y+z^2) \\  \\ =2xi-j+2zk

Also, the normal to the given plane 3x + 2y + 7z = 2 is given by:

3i+2j+7k

Equating the two normal vectors, we have:
</span>
2x=3\Rightarrow x= \frac{3}{2}  \\  \\ -1=2 \\ \\ 2z=7\Rightarrow z= \frac{7}{2}

Since, -1 = 2 is not possible, therefore there exist no such point <span>on the paraboloid y = x^2 + z^2 such that the tangent plane is parallel to the plane 3x + 2y + 7z = 2</span>.
4 0
3 years ago
Jean noticed that the tar pit bubbled more on warm days than on cool days or at night. Is the air temperature the dependent vari
andreev551 [17]
It's indepentent. In fact, the tar pit's temperature DEPENDS (dependent variable) by the air's temperature (independent variable).
6 0
3 years ago
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