What is the equation of a line that has the same slope as y=7-2x and the same y-intercept as y=-7x-5

Mathematics

1 answer:

wariber [46]2 years ago

3 0

Answer:

Hello!

First you will want to get both of equations into slope-intercept form. That way, you can easily determine the y-intercept of the first equation and the slope of the second equation.

When 2 lines have the same slope, that means they will be parallel.

y+4x-10=0 ; add 10 to each side

y+4x=10 ; subtract 4x from each side

y = -4x + 10 ; y-intercept = 10

1x + 7y = 6 ; subtract 1x from each side

7y = -1x + 6 ; divide both sides by 7

y = -1/7 x + 6/7 ; slope = -1/7

So, a line with the same y-intercept as the first, and same slope (parallel) as the second would have the equation:

y = -1/7x + 10

Step-by-step explanation:

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<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>

Generally, the equation for is mathematically given as

The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:

$A=\iint_{R_{x y}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y(1)$

The partial derivatives of a function are f x and f y.

$\begin{aligned}&Z=f(x)=6-3 x-2 y \\&=\frac{\partial f(x)}{\partial x}=-3 \\&=\frac{\partial f(y)}{\partial y}=-2\end{aligned}$

When these numbers are plugged into equation (1) and the integrals are given bounds, we get:

$&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\$

$&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}$

In conclusion, the area is

A=3 √4 units ^2