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13+x=20

20-13=x

20-13=7

Therefore x=7

Write 60^2 as the product of its prime factors

60^2

=(3*5*2*2)^2

=2^4*3^2*5^2

No,the max area is always a circle.(And the  the max volume enclosed by min surface area is always a sphere.)

Fence length  = 200 metres      which can enclose a circular area of

pi (r^2).                 So find r,the radius.

Circumference  = 200   =  2 pi(r)     

pi (r)                 =100

radius  is  100/(pi)         Area is  pi   X { (100)^2} /(pi^2) }

                                      Area = (100^2)  /pi      = 3183.098   which is greater than the area of the square.

The most area efficient shape is a circle.  The most volume efficient shape is a sphere.That's why planets are not cube shaped.

pythagoras theorem: how can you find the length of two sides of a triangle when only given one side

You can't, you need to be given two sides and a right angle to use pythagoras's theorum. But if you have another angle you might also be able to do it.  

13+x=20

13+x-13=20-13

x=7

The five circles making up this archery target have diameters of length 2,4,6,8, and 10.

What is the total red area?

Simplify your answer as much as you can.

Let A = area of a circle \(= \pi r^2\)
Ler r = radius of a circle

\(\begin{array}{|rclrcl|} \hline A_1 &=& \text{ area of 1st red circle and radius } & r_1 &=& 1 \\ &=& \pi \cdot 1^2 \\ &=& 1\pi \\ A_2 &=& \text{ area of 1st white circle and radius } & r_2 &=& 2 \\ &=& \pi \cdot 2^2 \\ &=& 4\pi \\ A_3 &=& \text{ area of 2nd red circle and radius } & r_3 &=& 3 \\ &=& \pi \cdot 3^2 \\ &=& 9\pi \\ A_4 &=& \text{ area of 2nd white circle and radius } & r_4 &=& 4 \\ &=& \pi \cdot 4^2 \\ &=& 16\pi \\ A_5 &=& \text{ area of 3rd red circle and radius } & r_5 &=& 5 \\ &=& \pi \cdot 5^2 \\ &=& 25\pi \\ \hline \end{array} \)

The total red area \(= A_\text{red}\)

\(\begin{array}{|rcll|} \hline A_\text{red} &=& A_5-A_4+A_3-A_2+A_1 \\ &=& 25\pi-16\pi+9\pi-4\pi+1\pi \\ &=& 15\pi \\ &=& 47.1238898038\ \text{units}^2 \\ \hline \end{array} \)

MY UNDERSTANDING (MIGHT BE WRONG)

Modulo is like the "remainder" of 2 positive numbers. For example, 4 mod 2 = 0 because 4/2 = 2 r.0 and 69 mod 21 = 6 because 69/21 = 3 r.6.

okay. You can put the index by doing \sqrt[index]{base}

If the side of the square is 12 cm, then the radius of the circle is half of that, or 6 cm. The circumference of a circle is 2*Pi*r. Substituting r, we get

\(2\Pi6cm=12\Pi\approx37.7cm\)

Thanks asinus :)

For this, we need to transorm the eq. to standard form:

\(x^2-6x=\frac{1}{2}x+42\)

\(x^2-\frac{13}{2}x-42=0\)

Now to use the quadratic formula:

\(a=1\)

\(b=-\frac{13}{2}\)

\(c=-42\)

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

\(x = {-(-\frac{13}{2}) \pm \sqrt{(\frac{13}{2})^2-4(1)(-42)} \over 2(1)}\)

\(x = \frac{\frac{13}{2} \pm \sqrt{210+{1 \over 4}}}{2}\)

\(x = {{13 \over 2}\pm(14+{1 \over 2})\over 2}\)

\(x = {10+{1\over2}, -4}\)

The polynomial function y = x^3 -3x^2 + 16x - 48 has only one non-repeated x-intercept.

What do you know about the complex zeros of the function?

Degree 3
The zero set of discriminant of the cubic  \({\textstyle ax^{3}+bx^{2}+cx+d\,} \) has discriminant
  \( {\textstyle b^{2}c^{2}-4ac^{3}-4b^{3}d-27a^{2}d^{2}+18abcd\,.}\)
The discriminant is zero if and only if at least two roots are equal.
If the coefficients are real numbers, and the discriminant is not zero,
the discriminant is positive if the roots are three distinct real numbers,
and negative if there is one real root and two complex conjugate roots.

\(\begin{array}{|rcll|} \hline && b^{2}c^{2}-4ac^{3}-4b^{3}d-27a^{2}d^{2}+18abcd \quad |\quad a=1\quad b= -3 \quad c = 16\quad d = -48 \\\\ &=& (-3)^{2}\cdot(16)^{2}-4\cdot 1 \cdot(16)^{3}-4\cdot(-3)^{3}\cdot(-48)-27\cdot 1^{2}\cdot(-48)^{2}+18\cdot 1 \cdot (-3) \cdot (16) \cdot (-48) \\ &=& 9\cdot 16^{2}-4\cdot 16^{3}-4\cdot (-27) \cdot (-48) -27\cdot(-48)^{2}+18\cdot 3 \cdot (16) \cdot 48 \\ &=& 9\cdot 256-4\cdot 4096-4\cdot 1296 -27\cdot 2304 + 41472 \\ &=& 2304 - 16384 - 5184 -62208 + 41472 \\ &=& -40000 \\ \hline \end{array}\)

discriminant  = -40000 < 0  there is one real root and two complex conjugate roots.

The max area is always a square:

P = 200, s = 200/4 = 50, A = 50*50 = 2500 square meters.

Part (1).  Not necessary to do all the above work. -tip ; always do a quick sketch to see that

When x is  negative and large , y is  negative and larger still.  When x is positive and large,y is positive and larger still.( The cubic term eventually outweighs all the others as x gets larger and larger.)

When x is  zero,  y =   -48.

so we know that  the function increases from negative infinity in the  3rd quadrant,crosses the y-axis at the point

( 0,-48)  and then keeps increasing. So it only has one real root.

A cubic has 3 roots,so the other two roots must be complex. (We can say more about these,but it's not asked for here.)

Part(2)

we can write a function as the product of some polynomial P,its quotient Q,and a remainder.

ie  y = P(x-2) + R   or f(x) = P(x-2) + R

so  4x^3-5x^2+3x-1  = P(x-2) + R

then f(2)  =R       = 17

How did you get the height asinus ?   Thank's Melody!

parallelogram \(ABCD\)

 \(\overline{AB}=a\\ \overline{BC}=s\\ triangle\ BAD=45°\)

\(perpendicular\ from\ D\ to\ a\ to\ point\ E\ is\ h.\)

\(AED\ is\ an\ isosceles\ right\ triangle.\)  \((\overline {DE}=h)=\overline{AE}\)

Set of Pythagoras:

\(\overline {AD}\ ^2=\overline{DE}\ ^2+\overline{AE}\ ^2\)

\(s ^2=2h^2\)

\(h^2=\frac{s^2}{2}\)

\(\large h=\frac{s}{\sqrt{2}}\)

  !

\(h^2+h^2=6^2\\ 2h^2=36\\ h^2=18\\ h=\sqrt{18}\\ h=3\sqrt2\)

So

\(area=base \times height\\ A=8\times 3\sqrt2\\ A=24\sqrt2\;\;units^2\\\)

How did you get the height asinus ?    

To answer the first queston, first factor the polynomial function.

\(y={x}^{3}-{3x}^{2}+16x-48\)

\(y=({x}^{3}-{3x}^{2})+(16x-48)\)

\(y={x}^{2}(x-3)+(16x-48)\)

\(y={x}^{2}(x-3)+16(x-3)\)

\(y=({x}^{2}+16)\times(x-3)\)

\(y=({x}^{2}+{4}^{2})\times(x-3)\)

\(y=(x-4)\times(x+4)\times(x-3)\)

Now set each x value to zero and solve

\(x-4=0\)

\(x-4+4=0+4\)

\(x-0=0+4\)

\(x=0+4\)

\(x=4\)

\(x+4=0\)

\(x+4-4=0-4\)

\(x+0=0-4\)

\(x=0-4\)

\(x=-4\)

\(x-3=0\)

\(x-3+3=0+3\)

\(x-0=0+3\)

\(x=0+3\)

\(x=3\)

Now plug each answer to the original polynomial function to see if the answers will fit in the original polynomial function.

\(y={x}^{3}-{3x}^{2}+16x-48\)

\(0={x}^{3}-{3x}^{2}+16x-48\)

\(0={4}^{3}-3({4}^{2})+16(4)-48\)

\(0=64-3({4}^{2})+16(4)-48\)

\(0=64-3(16)+16(4)-48\)

\(0=64-48+16(4)-48\)

\(0=64-48+64-48\)

\(0=16+64-48\)

\(0=80-48\)

\(0≠32\)

\(y={x}^{3}-{3x}^{2}+16x-48\)

\(0={x}^{3}-{3x}^{2}+16x-48\)

\(0={(-4)}^{3}-{3(-4)}^{2}+16(-4)-48\)

\(0=-64-{3(-4)}^{2}+16(-4)-48\)

\(0=-64-3(16)+16(-4)-48\)

\(0=-64-48+16(-4)-48\)

\(0=-64-48+(-64)-48\)

\(0=-112+(-64)-48\)

\(0=-176-48\)

\(0≠-224\)

\(y={x}^{3}-{3x}^{2}+16x-48\)

\(0={x}^{3}-{3x}^{2}+16x-48\)

\(0={3}^{3}-{3(3}^{2})+16(3)-48\)

\(0=27-{3(3}^{2})+16(3)-48\)

\(0=27-3(9)+16(3)-48\)

\(0=27-27+16(3)-48\)

\(0=27-27+48-48\)

\(0=0+48-48\)

\(0=48-48\)

\(0=0\)

The only real answer is x=3.  Becasue x=4 and x=-4 are not real answers, that means they are imaginary answers.  This means that B is the correct answer.

I know how to solve the second question but, because I do not know how to type it out on this forum, I will leave that to someone who does know how to answer the question and how to type it out on this forum.

Now for the next one:

\(\begin{array}{rcr} wxy&=&10 \\ wyz&=&5 \\ wxz&=&45 \\ xyz&=&12 \\ \end{array} \)
Find \( w+x+y+z\)

1.

\(\begin{array}{|rcll|} \hline xyz &=& 12 \\ xyz \cdot \frac{wyz}{wxy} \cdot \frac{wxz}{wxy} &=& 12 \cdot \frac{5}{10} \cdot \frac{45}{10} \\ xyz \cdot \frac{z}{x} \cdot \frac{z}{y} &=& 12 \cdot \frac{5}{10} \cdot \frac{45}{10} \\ z^3 &=& 12 \cdot \frac{1}{2} \cdot \frac{9}{2} \\ z^3 &=& 3\cdot 9 \\ z^3 &=& 3^3 \\ \mathbf{z} & \mathbf{=} & \mathbf{3} \\ \hline \end{array} \)

2.

\(\begin{array}{|rcll|} \hline && w+x+y+z \\ &=& \left( \frac{wxy}{xyz} + \frac{wxy}{wyz} + \frac{wxy}{wxz} \right) \cdot z + z \\ &=& \left( \frac{10}{12} + \frac{10}{5} + \frac{10}{45} \right)\cdot z + z \\ &=& \left( \frac{5}{6} + 2 + \frac{2}{9} \right)\cdot z + z \quad & | \quad z= 3 \\ &=& \left( \frac{5}{6} + 2 + \frac{2}{9} \right)\cdot 3 + 3 \\ &=& \frac{5}{2} + 6 + \frac{2}{3} + 3 \\ &=& 9 + \frac{5}{2} + \frac{2}{3} \\ &=& 9 + \frac{15+4}{6} \\ &=& 9 + \frac{19}{6} \\ &=& 12 + \frac{1}{6} \\ \hline \end{array}\)

Find the area of a parallelogram with sides of length 6 and 8, and with an interior angle with measure 45 degrees. (No trigonometry please!!!)

Basic line a = 8
Sidelines s = 6
Height \(h=\frac{s}{\sqrt{2}}\)

\(A=a\times h=a\times\frac{s}{\sqrt{2}}=8\times\frac{6}{\sqrt{2}}=\frac{48}{\sqrt{2}}\)

\(​A=33.94112549695..\)

  !

4 liters of water a day??!! OK!.

4 x 7 = 28 liters of water that Cheryl has to drink.

1 liter = 1,000 ml.

28 x 1,000 =28,000 ml. of water that Cheryl has to drink in 7 days!.

4 x 1,000 =4,000 ml. of water per day.

80,000 /4,000 = 20 days - time it will Cheryl to drink 80,000 ml. of water.

Nth Term = F x R^(N - 1), where F=First term, R=Common ratio, N=Number of terms

7 x R^(7 - 1) =5103, solve for R

7 x R^6 =5103  divide both sides by 7

R^6 =5103/7

R^6 = 729          take the 6th root of both sides

R =729^(1/6)

R = 3 Common ratio.

5th term =7 x 3^(5 - 1)

5th term =7 x 3^4

5th term =7 x 81

5th term = 567

thank you

See answer here:

http://web2.0calc.com/questions/metric-show-work_2