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3 years ago

In the equation y=mx,which statement is not true about m?

Mathematics

2 answers:

bearhunter [10]3 years ago

7 0

M is always positive

UkoKoshka [18]3 years ago

5 0

I'm pretty sure m is always positive

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Your Assignment

mote1985 [20]

Answer:

u b

Step-by-step explanation:

3 0

3 years ago

Or Simplify. Express your answer as a single term using exponents. 303*9x303*9

dmitriy555 [2]

${303}^{9} \times {303}^{9} \\ = {303}^{9 + 9} \\ = {303}^{18}$

Answer:

${303}^{18}$

Hope you could get an idea from here.

Doubt clarification - use comment section.

6 0

3 years ago

How do you find circumstance of a circle using the area? please help!

kotykmax [81]

Answer:

Step-by-step explanation:

The formula C = 2√πA is designed to find a circle's circumference using the area (A). Alternatively, you can solve the equation A = πR2 in reverse to find R, then plug R into the circumference equation.

3 0

3 years ago

Find the area of the part of the plane 3x 2y z = 6 that lies in the first octant.

gavmur [86]

The area of the part of the plane 3x 2y z = 6 that lies in the first octant is mathematically given as

A=3 √(4) units ^2

<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