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Your Question:

"can you take away plus four from minus three"

Solution:

If we are "taking away" something, then we are subtracting.

So -3 was the initial number that we want to pull from.  As set up, it would be:

-3 - 4 = X

We know in Algebra, there is no such thing as subtraction.  It is just a negative number.  So in reality, we have:

-3 + - 4 = X

With negative numbers, we really can't "take away" from them.  If you have -.03 cents in your pocket, but you spend another -.04, then you are even worse off, as you are now -.07 in debt.  To put this in back into the math problem:

-3 + - 4 = -7.

Now, if we want to actually "take away" from -3, we have to remember the number line.

< ---- (-3) --- (-2) --- (-1) --- ( 0 ) --- (+1) --- (+2) --- (+3) ---->

However, in order to this, we need the math problem to change to:

-3 + 4 = X

If we start at the -3 on the number line above, count over 4 spaces (because we have the positive four next to it), then we wind up at positive 1.

-3 + 4 = 1.

Using similar triangles twice, and  calling the length of the side of the square x,

in the triangle CYZ,  CY/x = 4/3, so CY = 4x/3,

in the triangle WXA,  x/XA = 3/4, so XA = 3x/4.

The hypotenuse of the triangle is

CY + YX + XA  = 4x/3 + x + 3x/4 = 37x/12 = 5,

so x = 60/37.

Irritating, Alan's solution appeared during the time I was typing mine in.

Using similar triangles twice, and  calling the length of the side of the square x,

in the triangle CYZ,  CY/x = 4/3, so CY = 4x/3,

in the triangle WXA,  x/XA = 3/4, so XA = 3x/4.

The hypotenuse of the triangle is

CY + YX + XA  = 4x/3 + x + 3x/4 = 37x/12 = 5,

so x = 60/37.

Hi Juriemagic,

I remember seeing this question before and I though I commented the first time round.

I think something is missing from your question.

You have twqo unknowns.

The first that you are trying to find is the regular 4 yearly deposit.

BUT

You know that the investment grows to 20,000 but you are not told how long ths will take. You need to know how long it will take to grow to 20,000 before you can work out a money value for the regular 4 yearly deposit.

Are you given a time anywhere?

"Right ABC has AB=3, BC=4, and AC=5. Square XYZW is inscribed in ABC with X and Y on AC, W on AB, and Z on BC. What is the side length of the square?"

Good day Melody!!,

There were no formulas given. I would imagine the formula A=P(1+i)^n must be used?...I just do not know how to interpret the question, so as to know where to start with this...I really hope you can help. Thank you kindly Melody..

Right ABC has AB=3, BC=4, and AC=5. Square XYZW is inscribed in ABC with X and Y on AC, W on AB, and Z on BC. What is the side length of the square?

\(\small{ \begin{array}{lcl} XW = WZ = x \qquad u = ZB \qquad v = WB\\ \end{array}\\ \begin{array}{lrcl} (1) & \frac{v}{x} &=& \frac{3}{5} \quad \rightarrow \quad v = \frac{3}{5} x\\ (2) & \frac{u}{x} &=& \frac{4}{5} \quad \rightarrow \quad u = \frac{4}{5} x\\\\ & \sin{(A)} &=& \frac{x}{3-v} \\ & \sin{(B)} = \sin{(90^{\circ}-A)} = \cos{(A)} &=& \frac{x}{4-u} \\ (3) & [\sin{(A)}]^2 + [\cos{(A)}]^2 =1 &=& \left(\frac{x}{3-v}\right)^2 + \left(\frac{x}{4-u}\right)^2 \\\\ \\ \hline \\ &\left( \dfrac{x}{3-v} \right)^2 + \left(\dfrac{x}{4-u} \right)^2 &=& 1 \\\\ & \dfrac{x^2}{ (3-v)^2 } + \dfrac{x^2}{ (4-u)^2 } &=& 1 \qquad v = \frac{3}{5} x \qquad u = \frac{4}{5} x\\\\ & \dfrac{x^2}{ \left(3-\dfrac{3}{5} x \right)^2 } + \dfrac{x^2}{ \left(4-\dfrac{4}{5} x \right)^2 } &=& 1 \\\\ & \dfrac{x^2}{ 3^2 \left(1-\dfrac{1}{5} x \right)^2 } + \dfrac{x^2}{ 4^2\left(1-\dfrac{1}{5} x \right)^2 } &=& 1 \\\\ & \dfrac{x^2}{ 3^2 } + \dfrac{x^2}{ 4^2 } &=& \left(1-\dfrac{1}{5} x \right)^2 \\\\ & \dfrac{x^2}{ 3^2 } + \dfrac{x^2}{ 4^2 } &=& \dfrac{ (5-x)^2 }{5^2}\\\\ & x^2 \left(\dfrac{1} { 3^2 }+ \dfrac{1}{ 4^2 } \right) &=& \dfrac{ (5-x)^2 }{5^2}\\\\ & x^2 \left(\dfrac{4^2+3^2} { 3^24^2 } \right) &=& \dfrac{ (5-x)^2 }{5^2} \qquad | \quad 3^2+4^2=5^2\\\\ & x^2 \left(\dfrac{5^2} { 3^24^2 } \right) &=& \dfrac{ (5-x)^2 }{5^2} \qquad | \quad 5^25^2=5^4\\\\ & x^2 \left( \dfrac{5^4} { 3^24^2 } \right) &=& (5-x)^2 \\\\ & x^2 \left( \dfrac{5^4} { 3^24^2 } \right) &=& 5^2-2\cdot 5 x + x^2 \\\\ & x^2 \left( \dfrac{5^4} { 3^24^2 } \right) -x^2 &=& 5^2-2\cdot 5 x \\\\ & x^2 \left( \dfrac{5^4} { 3^24^2 } -1 \right) &=& 5^2-2\cdot 5 x \\\\ & x^2 \left( \dfrac{5^4-3^24^2} { 3^24^2 } \right) &=& 5^2-2\cdot 5 x \\\\ & \mathbf{ x^2 \left( \dfrac{5^4-3^24^2} { 3^24^2 } \right) +2\cdot 5 x - 5^2 } & \mathbf{=}& \mathbf{ 0 }\\\\ \\ \hline \\ & x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \quad a= \frac{5^4-3^24^2} { 3^24^2 } \quad b=2\cdot 5 \quad c=-5^2 \\\\ & x^2 \left( \frac{5^4-3^24^2} { 3^24^2 } \right) +2\cdot 5 x - 5^2 &=& 0 \\\\ & x_{1,2} &=& {-2\cdot 5 \pm \sqrt{(-2\cdot 5)^2 - 4\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right) \cdot(-5^2) } \over 2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right)} \\\\ & x_{1,2} &=& {-2\cdot 5 \pm \sqrt{2^25^2 + 2^25^2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right) } \over 2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right)} \\\\ & x_{1,2} &=& {-2\cdot 5 \pm 2\cdot 5 \sqrt{1 + \left( \frac{5^4-3^24^2} { 3^24^2 }\right) } \over 2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right)} \\\\ & x_{1,2} &=& {-2\cdot 5 \pm 2\cdot 5 \sqrt{\frac{3^24^2+ 5^4-3^24^2} { 3^24^2 } } \over 2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right)} \\\\ & x_{1,2} &=& {-2\cdot 5 \pm 2\cdot 5 \sqrt{\frac{5^4} { 3^24^2 } } \over 2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right)} \\\\ & x_{1,2} &=& {-2\cdot 5 \pm 2\cdot 5 \left( \frac{5^2} { 3\cdot 4 } \right) \over 2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right)} \\\\ & x_{1,2} &=& {-10 \pm \frac{250} { 12 } \over 2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right)} \\\\ & x_{1,2} &=& {-10 \pm \frac{250} { 12 } \over \frac{962} { 144 } } \\\\ & x &=& {-10 + \frac{250} { 12 } \over \frac{962} { 144 } } \\\\ & x &=& {\frac{130} { 12 } \over \frac{962} { 144 } } \\\\ & x &=& \left( \frac{130} { 12 } \right) \cdot \left( \frac{ 144 }{962} \right) \\\\ & x &=& \left( \frac{130} { 12 } \right) \cdot \left( \frac{ 12^2 }{962} \right) \\\\ & x &=& \frac{130\cdot 12} { 962 } \\\\ & x &=& \frac{1560} { 962 } \\\\ & x &=& \frac{780} { 481 } \\\\ & x &=& \frac{60} { 37 } \\\\ & \mathbf{x} &\mathbf{=}& \mathbf{ 1.621\overline{621} } \end{array}\\ \text{The side length of the square is } \mathbf{1.621\overline{621}} }\)

Good one Mwashington18        LOL    

Let his mother's age 6 years ago  = x      and his age then was (1/5)x

And now we know that :

3 times his age now = his mother's age now....so........

3 [ (1/5)x + 6 ] =  x + 6    simplify

(3/5)x + 18 = x + 6     subtract (3/5)x , 6 from both sides

12  = (2/5)x      multiply both sides by 5/2

12 [ 5/2]  = x   = 60/2  = 30

This was his mother's age 6 years ago......so.....she must be 36, now

It is the same as 'asin'

Hi Juriemagic,

I assume you have been given formulas.

Have you been given a formula called

Future value of an ordinary annuity

as well as one called

Future value of an annuity due?

Maybe you have only been given the first one, or less likely, are you suppose to do it without either of these formula?

The easiest formula is future value of an annuity due but if you don't have it we can use future value of an ordinary annuity almost as easily.

Let me know, and perhaps enter the formula that you are given, and I will then continue.    

12.64

The only  difference is the volume of a cylinder  with  a radius and height of 2 cm  and the volume of a hemisphere with a radius of 2cm

It obvious that the cylinder will have a greater volume than the hemisphere.........and this difference is

pi (2)^2 * 2  -  (4/6)pi (2)^3   = pi [ 8 - 16/3]  = 8pi/3   cubic cm

So...the hemisphere uses  [8pi/3] cm^3   less ice cream

2/3 of 3/4   =  6/12  = 1/2 

There may be other methods of solving this one.....but here's my attempt.....

Locate  point D  on BC and let it lie x units from the origin......and let a be the side of the square

Locate point E on  AC  and  F on AB  such that  DE = DF  = a   and DE,DF are perpendicular to each other

Now angle ACB  = angle FDB = atan(3/4)

 So, cos [atan(3/4)] = DB/ FD = x / a    →  x = ( a)cos[ atan(3/4)]

And sin ACB = sin [atan (3/4)]

So, sin [atan (3/4) ]  = a / [ 4 - x]

And substituting for x, we have

sin [atan(3/4)] = a / [ 4 - a* cos [atan(3/4)]]

And with a little help from WolframAlpha,  a = 60/37.........and this is the side  of the square

And x = BD = (60/37)cos[atan(3/4)] = 48/37

Here's the [ approximate] pic :

How about: 1 - 1=0, or 0^1=0

Are you happy with that?

The record for the most amount of rain in the shortest amount of time in the US was 12 inches in 42 minutes. How much rain fell in 15 minutes? 28 minutes?

12/42=0.285714.....inches per minute.

15 X 0.285714=4.28571 inches in 15 minutes

28 X 4.28571  =8 inches in 28 minutes.

well if you think of the proportions, then you can take 81/18 = 4.5 where 4.5 is the ratio of the hem to the square inches. Then since they are proportionate(the dress of Rose and the doll), then you are able to use 4.5 as the dilation value, and thus multiply 40 (Rose's hemline) by (the dilation) 4.5 and that = 180 inches. You could also set up a proportion if that is easier to see.

If the volume of a cube is 5832 m3, then length of each side of the cube is  _____ metrecube

V=s^3

5832=s^3 take the cube of both sides

s=18 meters.

If you take trig......you will run across a "formula"  for the area of a sector of a circle

It's given by :

A = (1/2) r^2 (theta)      where "theta"  is the  "central angle"  of the sector measured  in radians  ....[ don't worry about the meaning of all these terms....... it isn't critical, here ]

The "central angle" measure of a sector formed by the whole circle  =  2*pi   radians

Thus,  the area  of the whole circle  =

(1/2) r^2  [ 2 * pi]  =

(1/2)(2)* pi * r^2  =

pi * r^2

Which is the familiar "formula" for the area of a circle......!!!!

Thus "pi" helps us find this area........

BTW.....before the "exact" value of "pi" was known [ not actually possible], mathematicians from India sometimes  used the value sqrt(10)  = about  3.16

The %error  here  is     [3.16 - 3.14] / 3.14 = about  0 .636%   [less than 1 percent !!!  ]

what is the area of a parallelogram with a base of 35/4 feet and a height of 16/3 feet in square feet

Area=35/4 X 16/3=560/12=46 2/3 square feet

"ton"    

Because Pi comes from a circle!!!

Pi=Circumference/Diameter(of a circle)

Two negatives ADDED together EQUALS a negative

Two Negatives MULTILIED equals a positive

Multiplying out the two terms....

(2sqrt3 + 4)(5sqrt3 - 2) =

10 (3) - 4 sqrt3 + 20sqrt3 - 8 =

30 + 16 sqrt3 -8 =

22 +16 sqrt3 =

22 + 16 (1.73205) = 49.713