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3 years ago

Is y = 7x +2 and x + 7y = 8 perpendicular, parallel, or neither ?

Mathematics

2 answers:

Vika [28.1K]3 years ago

5 0

Answer:

The lines are perpendicular.

Step-by-step explanation:

I graphed both of the equation on the graph below.

andrey2020 [161]3 years ago

3 0

Answer:

perpendicular

Step-by-step explanation:

You want them to be in the same form first (aka y=mx+b)

y=7x+2 is already in that form so you just need to change x+7y=8. First subtract x from both sides and then divide both sides by 7 to get y by itself:

x+7y=8

7y=-x +8

y= -x/7 + 8/7

To be able to tell if a line is perpendicular or not is to look at the slope: if the slope of one equation is the "opposite" version of the other, they are perpendicular.

The slopes are -x/7 and 7x so yes, they are perpendicular

** if they're parallel they will have the same slope and if they are neither they won't have a similar slope at all

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3 years ago

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Can you explain how a table can be used to find the rate of change

serg [7]

A table will generally give you an output value for each of several input values. To find the average rate of change over some range of inputs, divide the difference between output values by the difference between input values for the corresponding inputs.

For example, consider the table

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6 0

3 years ago

Find all points on the portion of the plane x+y+z=5 in the first octant at which f(x, y, z) = xy2z2 has a maximum value.

irina1246 [14]

Lagrange multipliers:

$L(x,y,z,\lambda)=xy^2z^2+\lambda(x+y+z-5)$

$L_x=y^2z^2+\lambda=0$
$L_y=2xyz^2+\lambda=0$
$L_z=2xy^2z+\lambda=0$
$L_\lambda=x+y+z-5=0$

$\lambda=-y^2z^2=-2xyz^2=-2xy^2z$

$-y^2z^2=-2xyz^2\implies y=2x$ (if $y,z\neq0$ )

$-y^2z^2=-2xy^2z\implies z=2x$ (if $y,z\neq0$ )

$-2xyz^2=-2xy^2z\implies z=y$ (if $x,y,z\neq0$ )

In the first octant, we assume $x,y,z>0$ , so we can ignore the caveats above. Now,

$x+y+z=5\iff x+2x+2x=5x=5\implies x=1\implies y=z=2$

so that the only critical point in the region of interest is (1, 2, 2), for which we get a maximum value of $f(1,2,2)=16$ .

We also need to check the boundary of the region, i.e. the intersection of $x+y+z=5$ with the three coordinate axes. But in each case, we would end up setting at least one of the variables to 0, which would force $f(x,y,z)=0$ , so the point we found is the only extremum.