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likoan [24]
3 years ago
7

Subtract 4 from n then multiply by 6

Mathematics
1 answer:
geniusboy [140]3 years ago
3 0

Answer:

multiply 4 and 6 and the answer should be n-24

Step-by-step explanation:

n-4*6

n-24

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20x^2/12x+8 simplified
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  1. 20/12 4
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I cant see the answers anyone else has the problem
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Answer: im having the same problem

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Is the GCF of two even numbers 2
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No, GCF means greatest common factor, meaning the largest number than can be divided into the number. 2 would be the lowest common factor or LCF.
EX.  200 & 440. In this case the GCF would be 40, but the LCF would be 2 because they are both even numbers


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4 years ago
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Find the vertex and focus of the<br> parabola.<br> y2 – 4y + 12x – 8 = 0
IceJOKER [234]

Answer:

1,2

-2 2

Step-by-step explanation:

6 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

Read more about the plane

brainly.com/question/1962726

#SPJ4

5 0
1 year ago
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