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3 years ago

WHAT ARE THE SIMILARITIES AND DIFFERENCES OF LINEAR AND NON-LINEAR SYSTEMS OF EQUATIONS.

1 answer:

3 0

SIMILARITIES
Some similarities are that for linear and non-linear systems of equations, you need the same number of equations as unknowns to solve the system and find the only solution for each variable.
DIFFERENCES
Some differences are
The linear system of equations has degree "1" in each of its variables.
The non-linear system of equations has degree "n" in each of its variables.
where n = 1,2,3, ..., i, ..., n.

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Can someone help on both of the questions, i dont get it

o-na [289]

Answer:

A. finished 1km

Step-by-step explanation:

b. 8km per hour

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3 years ago

You stand a known distance from the base of the tree, measure the angle of elevation the top of the tree to be 15â—¦ , and then

gogolik [260]

Answer:

The maximum possible error of in measurement of the angle is $d\theta_1 =(14.36p)^o$

Step-by-step explanation:

From the question we are told that

The angle of elevation is $\theta_1 = 15 ^o = \frac{\pi}{12}$

The height of the tree is h

The distance from the base is D

h is mathematically represented as

$h = D tan \theta$ Note : this evaluated using SOHCAHTOA i,e

$tan\theta = \frac{h}{D}$

Generally for small angles the series approximation of $tan \theta \ is$

$tan \theta = \theta + \frac{\theta ^3 }{3}$

So given that $\theta = 15 \ which \ is \ small$

$h = D (\theta + \frac{\theta^3}{3} )$

$dh = D (1 + \theta^2) d\theta$

=> $\frac{dh}{h} = \frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta$

Now from the question the relative error of height should be at most

$\pm p$ %

=> $\frac{dh}{h} = \pm p$

=> $\frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta = \pm p$

=> $d\theta = \pm \frac{\theta + \frac{\theta^3}{3} }{1+ \theta ^2} * \ p$

So for $\theta_1$

$d\theta_1 = \pm \frac{\theta_1 + \frac{\theta^3_1 }{3} }{1+ \theta_1 ^2} * \ p$

substituting values

$d [\frac{\pi}{12} ] = \pm \frac{[\frac{\pi}{12} ] + \frac{[\frac{\pi}{12} ]^3 }{3} }{1+ [\frac{\pi}{12} ] ^2} * \ p$

=> $d\theta_1 = 0.25 p$

Converting to degree

$d\theta_1 = (0.25* 57.29) p$

$d\theta_1 =(14.36p)^o$

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3 years ago

98 POINTS! PLEASE HELP! I been stuck on this assignment for over 2 weeks. My teacher gave me a simple explanation when I asked f

Colt1911 [192]

See the attached picture.

Create an equation for the volume of the box, find the zeroes, and sketch the graph of the function.

The resulting box has a volume

$V(x)=x(8-x)(12-x)=x^3-20x^2+96x$

because the volume of a box is the product of its width $(12-x)$ , length $(8-x)$ , and height $(x)$ .

find the zeroes

You know right away from the factored form of $V(x)$ that the zeroes are $x=0,8,12$ . (zero product property)

sketch the graph of the function

Easy to plot by hand. You know the zeroes, and you can check the sign of $V(x)$ for any values of $x$ between these zeros to get an idea of what the graph of $V(x)$ looks like. See the second attached picture.

Here's what I mean by "check the sign" in case you don't follow. We know $V(x)=0$ when $x=0$ and $x=8$ . So we pick some value of $x$ between them, say $x=1$ , and find that

$V(1)=1(8-1)(12-1)=7\cdot11=77$

which is positive, so $V(x)$ will be positive for any other $x$ between 0 and 8. Similarly we would find that $V(x)$ for $x$ between 8 and 12, and so on.

What is the size of the cutout he needs to make so that he can fit the most marbles in the box?

It's impossible to answer this without knowing the volume of each marble...

If Thomas wants a volume of 12 cubic inches, what size does the cutout need to be?

Thomas wants $V(x)=12$ , so you solve

$x^3-20x^2+96x=12$

While this is possible to do by hand, the procedure is tedious (look up "solving the cubic equation"). With a calculator, you'd find three approximate solutions

$x\approx0.1284$