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cricket20 [7]
3 years ago
6

Just do a c e g and i​

Mathematics
2 answers:
anygoal [31]3 years ago
4 0

Answer:

a) 0.47÷0.3 =1.567

c)0.068÷0.07 =0.971

e)0.53÷0.006=88.333

g)3.61÷1.1=3.282

I)6.56÷0.09=73.889

marta [7]3 years ago
3 0

Answer:

A, 1.57

C.0.971

E.88.3

G.3.28

I.72.9

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insens350 [35]
Rational: (-15.333), (sqt of 400), (8.62134), (15 5/6), (-3 4/5)

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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

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1 year ago
What is the inverse of y=-x-4
Nimfa-mama [501]

Answer:

f^-1 -1(x)=x+4

Step-by-step explanation:

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