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4

Answer:

Here are three methods...

Explanation:

Note that #6 = 2*3# has no square factors, so #sqrt(6)# cannot be simplified.

It is an irrational number somewhere between #sqrt(4) = 2# and #sqrt(9) = 3#

We can find rational approximations using any one of a number of methods (there are at least 25 different methods). I will present a few. In each of these examples I will use #sqrt(6) ~~ 5/2# as an initial approximation, since that lies between #2=sqrt(4)# and #3=sqrt(9)#.

Babylonian method

Given a rational approximation #p_i/q_i# to the square root of #n#, a better approximation is given by #p_(i+1)/q_(i+1)# where:

#{ (p_(i+1) = p_i^2+n q_i^2), (q_(i+1) = 2 p_i q_i) :}#

Using #n=6#, #p_0 = 5# and #q_0 = 2# we find:

#{ (p_1 = p_0^2+n q_0^2 = color(blue)(5)^2+color(blue)(6)(color(blue)(2)^2) = 25+24 = 49), (q_1 = 2 p_0 q_0 = 2(color(blue)(5))(color(blue)(2)) = 20) :}#

#{ (p_2 = p_1^2+n q_1^2 = color(blue)(49)^2 + color(blue)(6)(color(blue)(20)^2) = 2401+2400 = 4801), (q_2 = 2 p_1 q_1 = 2(color(blue)(49))(color(blue)(20)) = 1960) :}#

Given that we started with an efficient approximation (which we did), the number of correct significant digits of each successive approximation will be approximately the total number of digits in the numerator and denominator.

So:

#sqrt(6) ~~ 4801/1960 ~~ 2.4494898#

A calculator tells me:

#sqrt(6) ~~ 2.44948974278317809819#

So we were almost good to #8# significant digits as expected.

Generalised continued fraction

#sqrt(a^2+b) = a+b/(2a+b/(2a+b/(2a+b/(2a+...))))#

So putting #a=5/2# and #b=6-(5/2)^2 = 24/4-25/4 = -1/4# we find:

#sqrt(6) = 5/2-(1/4)/(5-(1/4)/(5-(1/4)/(5-(1/4)/(5-...))))#

We can truncate this early to find:

#sqrt(6) ~~ 5/2-(1/4)/(5-(1/4)/5)#

#color(white)(sqrt(6)) = 5/2-(1/4)/(99/20)#

#color(white)(sqrt(6)) = 5/2-5/99#

#color(white)(sqrt(6)) = 5/2-5/99#

#color(white)(sqrt(6)) ~~ 2.44949#

For more accuracy, use more terms.

Limiting ratio of integer sequence

Given that:

#2sqrt(6) ~~ 5#

Consider the quadratic with zeros #5+2sqrt(6)# and #5-2sqrt(6)#...

#(x-5-2sqrt(6))(x-5+2sqrt(6)) = (x-5)^2-4*6#

#color(white)((x-5-2sqrt(6))(x-5+2sqrt(6))) = x^2-10x+1#

So the zeros satisfy:

#x^2=10x-1#

Define a sequence recursively based on this quadratic:

#a_0 = 0#

#a_1 = 1#

#a_(n+2) = 10a_(n+1)-a_n#

Then the ratio between successive terms will tend to #5+2sqrt(6)#

The first few terms are:

#0, 1, 10, 99, 980, 9701, 96030, 950599#

Hence:

#sqrt(6) ~~ 1/2(950599/96030-5) = 1/2(470449/96030) = 470449/192060 ~~ 2.44948974279#

3

Answer:

$13,880

Explanation:

Since the question states the questioner doesn't understand math or finance, I'll start off with a brief explanation of what is going on in the hopes that the problem itself makes more sense.

Let's first talk depreciation and why it's important. Let's say a business buys a truck and it costs $35,000 (I know - it's the question we're working with... bear with me...). Let's also say we buy a packet of copy paper for $5 (and the business does a lot of photocopying and such and so will go through the copy paper quickly).

Here's the question - is it fair to treat the expense of the truck and the paper the same way? We're going to go through the copy paper quickly, but will we go through the truck as quickly? No - we won't. And so it isn't fair to treat the two the same and that is why we don't expense the truck in the year we buy it.

The truck, however, will reduce in value each year that it's used - wear and tear, parts getting used and getting older, etc. I'm sure you've seen a new truck before and also seen an old truck that's been used for many years - the two are worth very different amounts! Depreciation is an attempt to calculate the decreasing value of the truck year over year.

In most cases (unless the truck catches fire or something), even after many years of use, there is still a value to it - someone will pay something for that truck. That is the residual value and in this case is estimated to be $2,000.

And so our depreciation calculation is going to calculate the declining value of the truck from $35,000 when purchases new to five years later when it's estimated to be worth $2,000, or a decrease in value of $33,000.

There are many different ways to calculate the depreciation for each year. Many of them have higher depreciation amounts early in the truck's life and lower in later years (that first scratch on a new truck takes more value away than the 100th scratch where 99 are already there). The declining balance method is one way of doing that - we apply a method to the value of the truck at the end of each year.

The question is having us use a rate of "twice the straight-line rate". For a 5-year depreciation term, straight line depreciation would simply break the $33,000 into 5 chunks, or 20% of the value each year, and simply reduce the truck by that amount. Since we want double that, we'll use 40%. And we'll apply it to the value of the depreciable amount of truck each year. Like this:

#(("Year","Start value","dep. %","dep. amount","end value"),(1,33000,40%,13200,19800),(2,19800,40%,7920,11880))#

So at the end of year 2 there is $11,880 left of depreciable value in the truck. Keep in mind there is also the $2,000 residual value, and so the book value of the truck at the end of year 2 is:

#$11800+$2000=$13800#

And will be shown on the books this way:

#"Assets"#
#color(white)(00)"Truck"color(white)(0000000000000000000000)35,000#
#color(white)(00)"Accumulated Depreciation"color(white)(00)ul(-21,120)#
#color(white)(00)"Book Value"color(white)(00000000000000000)ul(13,880)#

2

Answer:

graph{(1/3)abs(x-3)+4 [-3.865, 11.935, -0.68, 7.22]}

Explanation:

Let's start with the function #y=x# and describe the transformations taken in order to make the current function.

#y=x#
graph{y=x [-10, 10, -5, 5]}

First, we're taking the absolute value, meaning every negative #y# value is flipped across the #x#-axis and made positive.

#y=abs(x)#

graph{y=absx [-10,10,-5,5]}

Now, we have the function in terms of:

#y=aabs(x-h)+k#

where #a=1/3#, #h=3#, #k=4#. So let me explain what each of these means.

The parameter #a# is being multiplied by the #x# values, which determines the slope of the lines, which is the #"rise"/"run"# of the function, or #(∆y)/(∆x)#. Since this is #1/3#, we know that for every one increase in y, we get three increases in x.

#y=1/3abs(x)#
graph{1/3absx}

Next, the h value determines how far right we shift the function. NOTE: By default, the value is negative. If there is. a plus sign, you shift this function left.

Since this value is 3, we shift this three units right.

#y=1/3abs(x-3)#
graph{y=1/3abs(x-3)}

Finally, we have the lonely #k# value, which just tells us how far we shift this up. This value is four, and therefore we shift the graph 4 units up.

#y=1/3abs(x-3)+4#
graph{y=1/3abs(x-3)+4 [-3,11,-1,8]}

1

Answer:

#x=5 and y=-1#

Explanation:

To solve a system using elimination, the co-efficients of one of the variable must be the same in both equations. It helps if they have opposite signs. Note that we have #+2x and -6x#

The #LCM# of #2 and 6# is #6#

#color(white)(xxxxxx)+2x+5y =5" "" "A#
#color(white)(xxxxxx)-6x+7y =-37" "B#

#A xx 3: rarr" "color(blue)(+6x)+15y =15" " " "C#
#color(white)(xxxxxxx.xx)color(blue)(-6x)+7y =-37" "B#

The #x# terms are now additive inverses. They add to #0#

#C+B: rarr color(white)(xxxxxx)22y =-22" "larr# no #x# term.

#color(white)(xxxxxxxxxxxxxxx) y = -1#

Substitute #-1# for #y# in any equation - let's use #A#

#color(white)(xxxxxx)2x+5(-1) =5" "" "A#
#color(white)(xxxxxxxx)2x" "-5 =5#
#color(white)(xxxxxxxxxxxxx)2x =10#
#color(white)(xxxxxxxxxxxxxx)x =5#

Check the values in another equation: Use #B#

#color(white)(xxxxxx)-6x+7y =-37" "B#
#color(white)(xxxxxx)-6(5)+7(-1) =-37#

#color(white)(xxxxxx)-30-7 =-37#
#color(white)(xxxxxxxxx)-37 =-37" "larr# the equation works out

1

Answer:

The tree is # 32 " feet tall" #.

Explanation:

Please refer to the attached diagram.
The measures will be in feet, but left-off until the final answer is found.

Given: A tree and a flagpole.
They are 20 feet apart.
The flagpole is 48 feet tall. The tree is shorter.
A shadow is cast and its end is 60 feet from the base of the flagpole and the end of the shadow is the same for the flagpole and the tree.

Find:
The height of the tree.

Assume:
1) The ground is level.
2) Both the flagpole and the tree are perpendicular to the ground, i.e. they make a #90^@# angle with the ground.
3) The flagpole, tree, and end of the shadow lie along the same straight line.

The diagram is labelled so that:
FG is the height of the flagpole;
TU is the height of the tree;
FS is the distance from the flagpole base to end of shadow.
TS is the distance from the tree base to end of shadow;

The two right triangles formed are similar, so the ratios of the parts are equal.
If we set-up the ratios with TS on top of one of the fractions, the algebra is easier.
# color(blue)( (TS)/(FS) = (TU)/(FG) )#

Calculate the length of TS.
# TS = FS - FT = 60 - 20 = color(red)40 #

Substitute the values into the proportion.
# (40)/(60) = (TU)/(48) #
# ( cancel(40) 2)/( cancel(60) 3) * 48 = TU #
# (2)/( cancel(3) 1) * cancel(48) 16 = TU #
# color(red)( 32 " feet" = TU ) #

Diagram:
enter image source here

1

Answer:

#"see explanation"#

Explanation:

#"the equation of a line in "color(blue)"slope-intercept form"# is.

#•color(white)(x)y=mx+b#

#"where m is the slope and b the y-intercept"#

#"a line with a slope of zero, however, is a special case"#

#"this indicates that the line is parallel to the x-axis passing"#
#"through all points in the plane with the same y-coordinate"#

#"the equation of this line is "y=c#

#"where c is the value of the y-coordinate"#

#"the point "(2,3)" has a y-coordinate of 3"#

#rArry=3larrcolor(blue)"is the equation of the line"#
graph{y-0.001x-3=0 [-10, 10, -5, 5]}

1

Answer:

#color(blue)((1/3)x+4 > y >= -x+1#

Explanation:

Given:

We are given the System of Inequalities:

#color(brown)(y < (1/3)x+4 # and

#color(brown)(y >= (-x)+1#

#color(green)(Step.1#

#color(brown)(y < (1/3)x+4 # - Image of the graph created using GeoGebra

enter image source here

#color(green)(Step.2#

#color(brown)(y >= (-x)+1# - Image of the graph created using GeoGebra

enter image source here

#color(green)(Step.3#

#color(brown)(y < (1/3)x+4 # and #color(brown)(y >= (-x)+1# - Image of the combined graphs created using GeoGebra

If you observe closely, you will find the solution in a visual form.

The solution to the system of inequalities is the darker shaded region, which is the overlap of the two individual regions.

enter image source here

#color(green)(Step.4#

If you want to view just the solutions, please refer to the image below:

enter image source here

Note:

If the inequality is < or >, graph of the equation has a dotted line.

If the inequality is ≤ or ≥, graph of the equation has a solid line.

This line divides the xy- plane into two regions: a region that satisfies the inequality, and a region that does not.

4

Answer:

desmos.com

Explanation:

Let's turn #f(x)=abs(x^2-2)# into a piece-wise function.

Now, we see that #absa=a# when #a>=0#
Similarly, we see that #absa=-a# when #a<0#

Therefore, we have:
#f(x)=x^2-2# when #x^2-2>=0#

#f(x)=-(x^2-2)# when #x^2-2<0#

Let's graph the two parabolas. When we do, we get:

desmos.com
Now, the blue parabola applies when #x^2-2>=0#, and green parabola applies when #x^2-2<0#.

Therefore, we have(focus on the red graph):
desmos.com
Now, to find its inverse, we have to reflect the red graph over the line #y=x#.

Let's try this mathematically.

Our piece-wise function was:
#f(x)=x^2-2# when #x^2-2>=0#

#f(x)=-(x^2-2)# when #x^2-2<0#

Let's find the inverse function for each situation.

Now, remember that #f(f^-1(x))=x# and #f^-1(f(x))=x#

Therefore, we can now find the inverse functions.

#f(x)=x^2-2=>x=(f^-1(x))^2-2#

=>#x+2=(f^-1(x))^2#

=>#+-sqrt (x+2)=f^-1(x)# when #x^2-2>=0#

Similarly,

#f(x)=-x^2+2=>x=-(f^-1(x))^2+2#

=>#x-2=-(f^-1(x))^2#

=>#2-x=(f^-1(x))^2#

=>#+-sqrt (2-x)=f^-1(x)# when #x^2-2<0#

We now graph these two sideways parabolas:
desmos.com
When #y>=sqrt2# or #y<= -sqrt2 # , then we apply our first function.

When #-sqrt2<y< sqrt2# , then we apply our second function.

We therefore have:

desmos.com

Just note that #+-sqrt(x+2)# and other relations like this one is not a function but a relation because there are more than one output for a given input.

2

Answer:

Please see below.

Explanation:

If we invest #$P# at #r%# per annum compounded #n# times in a year for #t# years then future value is #P(1+r/(nxx100))^(nt)#

(a) Here #P=7000#, #r=7.25%#, #t=8# years and as we are compounding annually i.e. #n=1#, then future value is

#7000(1+7.25/100)^8=7000(1+0.0725)^8#

= #7000xx1.0725^8=7000xx1.750566=$12253.96#

(b) Here #P=2900#, #r=9%#, #t=5# years and as we are compounding monthly i.e. #n=12#, then future value is

#2900(1+9/1200)^(5xx12)=2900(1+0.0075)^60#

= #2900xx1.0075^60=2900xx1.565681=$4540.47#

2

Answer:

#p(x)=1/24x^4-1/6x^3-2/3x^3+2/3x+2#

Explanation:

#"given the roots of a polynomial, say"#

#x=a,x=b" and "x=c#

#"then the factors are"(x-a),(x-b)" and "(x-c)#

#"the polynomial is the product of the factors"#

#rArrp(x)=k(x-a)(x-b)(x-c)larrcolor(blue)"k is a multiplier"#

#"identify the roots from the graph"#

#x=-2("multiplicity 2"),x=2" and "x=6#

#"factors are expressed as"#

#(x+2)^2,(x-2)" and "(x-6)#

#rArrp(x)=k(x+2)^2(x-2)(x-6)#

#"to find k use any other point on the graph"#

#"using the point "(0,2)" substitute into the equation"#

#2=k(4)(-2)(-6)=48krArrk=2/48=1/24#

#rArrp(x)=1/24(x+2)^2(x-2)(x-6)#

#color(white)(rArrp(x))=1/24(x^4-4x^3-16x^2+16x+48)#

#color(white)(rArrp(x))=1/24x^4-1/6x^3-2/3x^2+2/3x+2#
graph{1/24x^4-1/6x^3-2/3x^2+2/3x+2 [-10, 10, -5, 5]}