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ΔAXC   is similar to  ΔAYB  by  AA congruency

So

XC /AC = YB / AB

XC / 70  = 60 / 75 

XC / 70  =  4 / 5       

XC  =  (4 / 5)  * 70  =  56

thx your the best

1) To find a perpendicular line, the slope is the negative reciprocal.

Get the equation into y=mx+b form.

5x+8y=-9

Subtract 5x from both sides.

\(8y=-5x-9\)

Divide both sides by 8.

\(y=-\frac{5}{8}x-\frac{9}{8}\)

Take the negative reciprocal of m and use it with (10,10) in point-slope form, y-y₁=m(x-x₁).

\(y-10=\frac{8}{5}(x-10)\)

Multiply the 8/5 through and add 10 to both sides.

\(y=\frac{8}{5}x-16+10\)

\(y=\frac{8}{5}x-6\)

Then, m+b would be 8/5-6. This is 1.6-6, or -5.6.

2) The slope of a line is \(\frac{y_2-y_1}{x_2-x_1}\).

Plug in the points (1,2) and (7,4)

\(\frac{4-2}{7-1}=\frac{2}{6}=\frac{1}{3}\)

So the slope is 1/3. Now use point-slope form.

\(y-2=\frac{1}{3}(x-1)\)

Distribute 1/3 and add 2 to both sides.

\(y={1\over3}x-\frac{1}{3}+2\)

Combine.

\(y=\frac{1}{3}x-\frac{1}{3}+\frac{6}{3}\)

\(y=\frac{1}{3}x+\frac{5}{3}\)

3) Put 3y+2x=7 into y=mx+b form.

\(3y=-2x+7\)

Divide by 3.

\(y=-\frac{2}{3}+\frac{7}{3}\)

The slope of a line perpendicular to another is the negative reciprocal of the first slope.

\(m=\frac{3}{2}\)

So 3/2 is the answer to #3.

Very nice,  AT  !!!

Area of square  = 1 unit^2

The circle has a radius of  (1/2)

So...its area is :    pi   (1/2)^2  =  pi / 4  units^2

So    area of circle to area of square  =   [ pi / 4] : 1  =   pi / 4  ≈ .785  ≈  79%

x^4 + 5x^2 + 6  = 0

We  can factor this as

(x^2  + 3) (x^2 + 2)  = 0

Setting each factor to 0  we have that

x^2  + 3  =  0               x^2  + 2  = 0

x^2  =  -3                     x^2  =  -2

Note that no real numbers make these true, so, there are 4 complex zeroes

hallelujah! thx so much. Praise the Almighty God.

1) f(x)= x⁴+5x³-x²-5x

Statements 1, 3, and 4 are correct. 2 is not because f(5) does not equal 0.

The Rational Root Theorem is:\

Let f(x) be a polynomial with integral coefficients. The only possible rational zeros of f(x) are: 

\(\frac{p}{q}\)

where p is a divisor of the constant term and q is a divisor of the leading coefficient.

So, fators of the constant on top and factors of the LC on bottom.

\(\frac{\pm1, \pm2, \pm4}{\pm1,\pm2}\)

Now, pair them together.

Divide all of the top factors by 1 first, then 2.

\(\pm1,\pm2,\pm4,\pm \frac{1}{2}, \pm1,\pm2\)

Repeating factors can be dropped so the possible factors are \(\pm1,\pm2,\pm4,\pm \frac{1}{2}\) or \((x+1),(x-1),(x+2),(x-2),(x+4),(x-4),(2x+1),(2x-1)\).

So your possible factors are correct.

To see which ones are factors, I woud graph it.

This graph shows that \((-2,0), (-\frac{1}{2},0), (2,0)\) are the zeros. (Here is the graph: https://www.desmos.com/calculator/fx9umcmzu0)

As factors those are (x+2), (2x+1), and (x-2).

Assuming that your picture means that the diameter of the smaller square is the radius of the bigger square, here is the solution. 

Lets say that the radius of the small circle is x, hence the radius of the larger circle is 2x. Using the area of circles formula: a = πr^2, where r is the radius, the area of the smaller circle is πx^2 and the area of larger circle is 4πx^2

The πx^2 cancels out, leaving us with 1:4.

The simpler solution is that if two shapes are similar, the ratio of their area will be one of the corresponding side's square. In this case, the radius. 

sorry, my computer glitches, how do you delete a post?

http://www.mathsisfun.com/puzzles/5-pirates-solution.html

http://www.mathsisfun.com/puzzles/5-pirates-solution.html

http://www.mathsisfun.com/puzzles/5-pirates-solution.html

http://www.mathsisfun.com/puzzles/5-pirates-solution.html

http://www.mathsisfun.com/puzzles/5-pirates-solution.html

ths so much you are amazing.

The formula to calculate the area of a circle is π*r^2, where r is the radius and pi is 3.14159265358979......

With that said, we could calculate the area of the outer red ring, but we just know the diameter. Since the radius is half the length of the diameter, the area of the ring will be 25π - (8/2)^2π , since the red ring is the different of the two circles.

Using the same logic, the red ring smaller than that is 5π, and the red circle in the middle is 1π. Adding those numbers together, we will get

9π + 5π + π 

The answer to this problem is 15π. 

I hope this helped,

Supermanaccz

Whatever the highest degree is in a polynomial is the number of complex factors.

Therefore, in this equation there are 4 real complex numbers.

1) For the first, the number of complex zeros you have is always the same as your highest degree.

So, for that one it would be 4 complex zeros.

2) f(x)=x⁴+3x²−4

Here is the graph: https://www.desmos.com/calculator/zm7f4iwvrb

From the graph, the two zeros are (-1,0) and (1,0).

Now, I'm going to use long division for these.

Pull off all the coefficients (Don't forget the missing terms. This problem is really x⁴+0x³+3x²+0x-4)

I'm gong to use (-1,0) first, since it makes x+1 as a factor.

(Sorry if the picture is a bit fuzzy, I tried my hardest to get a clear one.)

(x+1)(x-1)(x²+4)

Now plug the x^2+4 into the quadratic formula.

In x²+4, a=1, b=0, and c=4.

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}={-0 \pm \sqrt{0^2-4(1)(4)} \over 2(1)}\)

Now simplify. -0 can just go away and so can 0^2.

\(x={ \pm \sqrt{-4(1)(4)} \over 2}\)

4*1*-4=-16

\(x={\pm \sqrt{-16} \over 2}\)

The square root of 16 is 4 and the square root of -1 is i.

\(x=\frac{\pm4i}{2}\)

This can be simplified to \(x=\pm2i\).

So, the factors are \((x+1)(x-1)(x+2i)(x-2i)\).

This question is impossible to answer, because we have no relation between the number of weeks and the number of candy bars sold. We could sell one in 30 weeks are 100 in 1 week, if you can post the original question, I would glady do it. 

Thanks,

Supermanaccz

When a graph just "kisses" the x axis, we will have 2n zeroes at that point

So...let's assume that  x = 2  we have two zeroes

Thus,  let us guess that   (x - 2)^2  =  x^2 - 4x + 4   divides the polynomial

                           3x^2   +  x     +  2   

  x^2 - 4x + 4   [  3x^4  -11x^3  +  10x^2  -  4x + 8  ]

                           3x^4  -12x^3  +  12x^2

                           ______________________________

                                      x^3    -      2x^2     -    4x   +  8

                                      x^3    -      4x^2     +   4x

                                     ________________________

                                                        2x^2   -     8x  + 8

                                                        2x^2   -     8x  + 8 

                                                        ________________

So...the factorization is   ( 3x^2   +  x  +  2  )  ( x - 2) ^2

It was easy even though I've barely done it before

Or in polynomial form it would be \((x-2)(2x^2-x-2)+3\).

Correct....!!!!

x=   -3, -2, -1, 0, 1,  2, 3, 

f(x)=63, 8, -1, 0, -1, 8, 63

Notice that when the graph progresses from (- 2, 8) to (-1, -1)...it passes through the x axis for the first time....so this is one zero

And then (-1, -1)  to (0,0)  ......given the fact that we're not told that graph reaches a posiitive value before it progresses to (1, -1)....we have to assume that it only "touches" the origin....this means that, -- at a minimum- two more "zeroes" are added

And from (-1, -1)  to (2,8) it passes through the x axis once more....and a last zero is added

So...at a minimum,  the graph has a degree of 4  [ the fact that the graph just "kissses" the origin means that the actual degree could also be 6, 8, 10, etc.   When a graph  just touches the x axis, it means that 2n zeroes are added.......

Your second one is correct !!!