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math and science from a professional academic scientist

Bossier City, LA (71111)

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Kerrie's Responses in WyzAnt Answers

how can you make math seem fun when you are trying to each someone?

I am 13 and i am trying to help my friend out with her math because she is teaching me art.

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Asked by T from Olathe, KS
21

Kerrie's Answer:

If your friend likes art, you can use that to motivate discussions about math.  For instance symmetry is a concept that has a lot of art application, as do triangles, squares, etc.  It all depends on what subject she is studying.  You can find interesting lesson plans and tools on teacher web sites, which are full of suggestions about how to motivate students.  Try the teacher channel, for instance.

Physics' Quesion

Pulling a Crate A worker drags a crate across a factory floor by pulling on a rope tied to the crate (Fig. 6-60). The worker exerts a force of 450 N on the rope, which is inclined at 38° to the horizontal, and the floor exerts a horizontal force of 125 N that opposes the motion.

Figure 6-60 (image is in this link http://www.webassign.net/hrw/05_44.gif )

(a) Calculate the magnitude of the acceleration of the crate if its mass is 315 kg.

(b) Calculate the acceleration if the crate's weight is 315 N.

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Asked by Taliyah from Riverdale, IL
10

Kerrie's Answer:

You must draw a force diagram on the box.

• The weight acts down and is perpendicular to the movement.  It doesn't do any "work"
• The Floor acts horizontally, opposing motion
• Normally, I would consider the friction to be proportional to the Normal force and do some calculations to figure out the normal force, but here they just tell you that there is opposition to motion that acts horizontally, so let's go with that...
• The man acts at an angle, but only the portion of the force he exerts horizontally does any "work" to make the crate move...so we have to figure out the horizontal part of his force.
• A triangle (vector) diagram would be set up so that the man's force is the hypothenuse, and the horizontal component is adjacent to the angle (38 degrees).  The vertical component is the opposite leg.
• If you remember your triangle trigonometry, this means the the magnitude of the horizontal vector is F*(cos θ)

Our balanced force equation is then:

M*a = F*(cos θ) - f,

• where f is the resistance from the floor

(a)         a = {F*(cos θ) - f} / M

={450*(0.788)-125}/315

=0.729 meters/s2

(b)   if the weight is 315 N, then the mass is 315 = M*(9.81)

solve for M = 32.11kg

and plug this in to the equation

a = {450*(0.788)-125}/32.1 = 7.15 m/s2

How do you write a piecewise function from a word problem

It costs 7.57 plus 8.77 per kilowatt hour for the first 400 and 6.574 per kilowatt usage over 400 in a month.

What is the cost of 300 in a month? 2638.57 how much for 700 in one month? 5487.77 What I need is a piecewise function for this x=Kwhr used a month and c= cost of usage

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Asked by Adin from Asheville, NC
00

Kerrie's Answer:

You are right, you need a piecewise function, which you can separate based on the the segments of the domain that are given in words...

First, there is the the segment from 0-400 kwh, then the segment from 400 - infinity (what an electric bill!)

Here is how to write it:

7.57 + 8.77x  ,     0<= x < 400 kilowatt hours

f(x) =            (5.57 + 877*400) + 6.574(x - 400),     x > 400 kilowatt hours

Imagine a great big "curly brace" to the left of the equal sign, that wraps onto the two different functions.

The trick is to realize that the first 400 kwh translate into a constant that is added into rate*time of the amount over.  You also have to correct x in the second equation, since you don't want to charge twice  for the first 400 hours

write an equation for the line that is perpendicular to the given line and that passes through the given point y=-1/3x+2;(4,2)

write an equation for the line that is perpendicular to the given line and that passes through the given point

Y=-1/3x+2;  (4,2)

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Asked by Shawn from New Albany, IN
00

Kerrie's Answer:

The slopes of two perpendicular lines, when multiplied, have a product of -1.  Therefore, given the eq of the line with a slope of -1/3, we can conclude that the perpendicular line will have a slope of 3.

so to start, our equation of interest is y=3x + b, where b is unknown.

Since we know that it passes through the point (4,2) we can plug the values of x and y into this equation to solve for b.  2 = 3*4 + b

2=12 + b

-10 = b

This gives the entire equation we are after:  y=3x-10.

is there a easier way to learnt Quadratic Equations?

i do not understand this topic and i need help

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Asked by Keonja from Macon, GA
00

Kerrie's Answer:

I think you should play angry birds! No really, we do it in math class all the time.  A quadratic equation describes a parabola, which just happens to be the shape of the curve made by the bird as it is sling-shotted through the air.  Well, really it is an inverted parabola, and I will get to that in a minute.

Here are quadratics in a nutshell:

1. The standard equation lists the terms in descending order with highest power first.  Ax2 + Bx + C.
• This form is not particularly useful, except that the sign of A will tell you whether the curve is right side up (concave up) or upside down (think "angry bird in flight") (concave down).  If A > 0 the parabola is right side up....
• Oh, and the constant C gives you the y- intercept, just like the constant does in the equation for a line
2. The intercept form of the equation A(x-p)(x-q).  This is the "factored" form of the equation, and reveals the only reason why any self respecting angry birds fan would ever want to factor a quadratic in the first place.  Because they want to find the x-intercepts, or in other words, where the angry bird will land!
• Since The "solution", "root" or "zero" of this equation is given by setting the whole thing to zero, we can use the zero multiplicity rule (ZMP) to find out where the function crosses the x axis.  The ZMP just says that when you multiply three unknown numbers and you know the product is zero, that one or more of the unknowns must be equal to zero.  So either
• A=0 (boring)
• (x-p) = 0 (possibly where the angry bird left the sling shot)
• (x-q) = 0 (where the angry bird will land.  Hopefully on those annoying pigs!)
3. Lastly, there is the vertex form of the equation (x-h)2 +k.  Once again, why would any serious sling-shotter be interested in this algebraic form of torture?  Well, this equation is useful for telling us the  highest point in flight (h,k) of the parabolic bird in question.  Putting the equation into this form is far more tortuous than the other bits of algebraic manipulation, and is called "completing the square".  In this short, angry treatise, it is beyond the scope of my sense of humor.  But fear not, there are internet resources which abound...if you are really interested, you can respond to my answer and I will continue....

Happy sling-shotting!

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