All Answers 358187 Answers

(x sqrt x^3)^4

= (sqrt(x^5)^4

= (x^(5/2))^4

= x^20/2 = x^10

(17y+2) / 5  - (5-3y)/3      "cross" multiply

(17y+2)(3)/(5*3) - (5-3y)(5)/(5*3)

= (51y+6 -25+15y)/15

= (66y - 19)/15

101     132   156   187       202     231    257    286

110     134   165   189       211     235     264    297

112     143   167   198       213     242     268

121     145   176                220     246     275                 

123     154   178                224     253     279   

Total = 35

F = 9/5 C + 32     SInce we want F = C   sub this in

F = 9/5 F + 32

- 4/5 F = 32

F = -5/4 (32) = -5/4*32 = -40

A lot of people just happen to know this:   -40 C = -40 F

It's not anything you are doing incorrectly.......Web2.0Calc has had a problem the last few days and will not let ANYone upload images.....try again later is all you can do.....or see posting from yesterday on how to post a website with your image using other software    

For rational solutions b^2-4ac must be a perfect square

\(b^2-4ac=20^2-4k^2\\ b^2-4ac=400-4k^2\\ b^2-4ac=4(100-k^2)\\ \)

so 100 - k^2 must be a perfect square

k=6, 8  are the only 2 that work.

100-36=64

100-64=36

 OSL4#13

\(\text{Let the number of nickels $=n$, $1$ nickel $= 5 $ cent } \\ \text{Let the number of dimes $=d$, $1$ dime $= 10 $ cent } \\ \text{Let the number of quaters $=q$, $1$ quater $= 25$ cent }\)

\(\begin{array}{|lrcll|} \hline (1) & \mathbf{ n+d+q } &=& \mathbf{30} \\\\ & 5n+10d+25q &=& 500 \qquad \text{in cents}\quad &| \quad : 5 \\ (2) & \mathbf{ n+2d+ 5q } &=& \mathbf{100} \\ \hline (2)-(1): & n+2d+ 5q -(n+d+q) &=& 100-30 \\ & n+2d+ 5q -n-d-q &=& 70 \\ & d+ 4q &=& 70 \quad &| \quad -4q \\ (3)& \mathbf{ d } &=& \mathbf{70- 4q} \\ \hline & n+d+q &=& 30 \\ &n&=& 30-d-q \quad &| \quad d =70- 4q \\ &n&=& 30-(70- 4q)-q \\ &n&=& 30-70+ 4q-q \\ (4)& \mathbf{ n } &=& \mathbf{-40+ 3q} \\ \hline \end{array} \)

\(\mathbf{d \geq 0}\)

\(\begin{array}{|rcll|} \hline 70- 4q &\geq& 0\quad &| \quad + 4q \\ 70 &\geq& 4q \quad &| \quad : 4 \\ \dfrac{70}{4} &\geq& q \\ q &\leq& \dfrac{70}{4} \\ q &\leq& 17.50 \\ \mathbf{q} &\leq& \mathbf{17} \\ \hline \end{array}\)

\(\mathbf{n \geq 0}\)

\(\begin{array}{|rcll|} \hline -40+ 3q &\geq& 0 \quad &| \quad +40 \\ 3q &\geq& 40 \quad &| \quad : 3 \\ q &\geq& \dfrac{40}{3} \\ q &\geq& 13.3333333333 \\ \mathbf{q} &\geq& \mathbf{14} \\ \hline \end{array} \)

\(\mathbf{14\leq q \leq 17 } \qquad q=\{14,\ 15,\ 16,\ 17 \} \)

There are 4 different combinations.

The combinations:

\(\begin{array}{|c|c|c|} \hline \text{q (Quaters)} & d=70- 4q \text{ (dimes)} & n=3q-40 \text{ (nickels)} \\ \hline 14 & 14 & 2 \\ 15 & 10 & 5 \\ 16 & 6 & 8 \\ 17 & 2 &11 \\ \hline \end{array}\)

Thank you kindly Alan!!..Have a blessed day!

First find the sides of the square:

\(\sqrt{64} \\Then\ you\ should\ use\ the\ Pythagorean\ theorem\ to\ get\\ \\\sqrt{64+64} \\\sqrt{128} \\8\sqrt{2}\)

Thats your diagonal:

8sqrt2

If

\(x+y=\dfrac{7}{12} \)

and

\(x-y=\dfrac{1}{12}\),

what is the value of \(x^2-y^2\)?
Express your answer as a common fraction.

\(\begin{array}{|rcll|} \hline (x-y)(x+y) &=& \dfrac{7}{12}\times \dfrac{1}{12} \quad | \quad (x-y)(x+y) = x^2-y^2 \\ \mathbf{x^2-y^2} &=& \dfrac{7}{12}\times \dfrac{1}{12} \\ &=&\mathbf{ \dfrac{7}{144}} \\ \hline \end{array} \)

What is the nearest integer to \((5+2\sqrt7)^4\) ?

\(\begin{array}{|rcll|} \hline && (5+2\sqrt7)^4 \\ &=& \left(~(5+2\sqrt7)^2~\right)^2 \\ &=& \left(~25+20\sqrt7+4\cdot 7~\right)^2 \\ &=& \left(~53+20\sqrt7~\right)^2 \\ &=& \left(~53^2+2\cdot 53\cdot 20\sqrt7 + 400\cdot 7 ~\right)^2 \\ &=& 5609+2120\sqrt7 \\ &=& 5609+2120\cdot 2.64575131106 \\ &=& 5609+5608.99277946 \\ &=& 5609+5609 \quad | \quad \text{to the nearest integer } \\ &=& \mathbf{11218} \\ \hline \end{array}\)

Yes, it is perfectly possible!

nvm. Sorry!

Nice Heureka! You beat me to it though... :P

A survey of 120 teachers determined the following:

70 had high blood pressure
40 had heart trouble
20 had both high blood pressure and heart trouble

What percent of the teachers surveyed had neither high blood pressure nor heart trouble?

Frist solution:

\(\begin{array}{|l|c|c|c|} \hline & \text{blood} & \overline{\text{blood}} \\ & \text{pressure} & \overline{\text{pressure}} \\ \hline \text{heart} & \mathbf{20} & 40-20=20 & \mathbf{40} \\ \text{trouble} & & & \\ \hline \overline{\text{heart}} & & \color{red}x =50-20 = \mathbf{30} \\ \overline{\text{trouble}} & \\ \hline & \mathbf{70} & 120-70=50 & \mathbf{120} \\ \hline \end{array}\)

Second solution:

\(\begin{array}{|l|c|c|c|} \hline & \text{blood} & \overline{\text{blood}} \\ & \text{pressure} & \overline{\text{pressure}} \\ \hline \text{heart} & \mathbf{20} & & \mathbf{40} \\ \text{trouble} & & & \\ \hline \overline{\text{heart}} & 70-20=50 & \color{red}x = 80-50 =30 & 120-40 = 80 \\ \overline{\text{trouble}} & \\ \hline & \mathbf{70} & & \mathbf{120} \\ \hline \end{array}\)

Traviso, your comment is slightly disturbing to me.

I am after all, a "young girl."

yep, lol

In the complex plane, what is the length of the diagonal of the square with vertices \(4,\ 3+5i,\ -2+4i, \text{ and } -1-i \) ?

\(\begin{array}{|rcll|} \hline && |(3+5i)-(-1-i)| \\ &=& |3+5i+1+i| \\ &=& |4+6i| \\ &=& 2|2+3i| \\ &=& 2\sqrt{2^2+3^2} \\ &=& \mathbf{2\sqrt{13}} \\ \hline \end{array}\)

or

\(\begin{array}{|rcll|} \hline && |(-2+4i)-(4)| \\ &=& |-2+4i -4| \\ &=& |-6+4i| \\ &=& 2|-3+2i| \\ &=& 2\sqrt{(-3)^2+2^2} \\ &=& \mathbf{2\sqrt{13}} \\ \hline \end{array}\)

Hi Alan,

so it is possible that both volume and surface area can be the same figure?...except of course for the square and cubic results?

Suppose \(x\), \(y\), and \(z\) form a geometric sequence.
If you know that and \(x + y + z = 18\),
find the value of \(x^2+y^2+z^2=612\).

I assume:

Geometric sequence:
\(x = a \\ y = ar \\ z = ar^2\)

\(\begin{array}{|lrcll|} \hline & \mathbf{x+y+z} &=& \mathbf{18} \\ &a+ar+ar^2 &=& 18 \\ &a(1+r+r^2) &=& 18 \quad |\quad 1+r+r^2 = \dfrac{1-r^3}{1-r} \qquad {\displaystyle |r|<1} \\\\ (1) &\mathbf{a\left(\dfrac{1-r^3}{1-r}\right)} &=& \mathbf{18} \qquad \text{or} \quad \mathbf{a} = \mathbf{18\left(\dfrac{1-r}{1-r^3}\right)} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline & \mathbf{x^2+y^2+z^2} &=& \mathbf{612} \\ &a^2+a^2r^2+a^2r^4 &=& 612 \\ &a^2(1+r^2+r^4) &=& 612 \quad &|\quad 1+r^2+r^4 = \dfrac{1-r^6}{1-r^2} \qquad {\displaystyle |r|<1} \\ (2)&\mathbf{a^2\left(\dfrac{1-r^6}{1-r^2}\right)} &=& \mathbf{612} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \dfrac{(2)}{(1)}: & \dfrac{ a^2\left(\dfrac{1-r^6}{1-r^2}\right) } { a\left(\dfrac{1-r^3}{1-r}\right)} &=& \dfrac{612}{18} \\\\ & \dfrac{ a\left(\dfrac{1-r^6}{1-r^2}\right) } { \left(\dfrac{1-r^3}{1-r}\right)} &=& 34 \\\\ & a \dfrac{(1-r^6)(1-r)}{(1-r^3)(1-r^2)} &=& 34 \quad |\quad 1-r^6 = (1-r^3)(1+r^3) \\\\ & a \dfrac{(1-r^3)(1+r^3)(1-r)}{(1-r^3)(1-r^2)} &=& 34 \quad |\quad 1-r^2 = (1-r)(1+r) \\\\ & a \dfrac{(1-r^3)(1+r^3)(1-r)}{(1-r^3)(1-r)(1+r)} &=& 34 \\\\ (3)&\mathbf{a\left(\dfrac{1+r^3}{1+r}\right)} &=& \mathbf{34} \qquad \text{or} \quad \mathbf{a} = \mathbf{34\left(\dfrac{1+r}{1+r^3}\right)} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1)=(3): & a=18\left(\dfrac{1-r}{1-r^3}\right) &=& 34\left(\dfrac{1+r}{1+r^3}\right) \\\\ & 18\left(\dfrac{1-r}{1-r^3}\right) &=& 34\left(\dfrac{1+r}{1+r^3}\right) \\\\ & 18(1-r)(1+r^3) &=& 34(1+r)(1-r^3) \quad & | \quad : 2 \\ & 9(1-r)(1+r^3) &=& 17(1+r)(1-r^3) \\ & 9(1-r+r^3-r^4) &=& 17(1+r-r^3-r^4) \\ & 9-9r+9r^3-9r^4 &=& 17+17r-17r^3-17r^4 \\ & 8r^4+26r^3-26r-8 &=& 0 \quad & | \quad : 2 \\ & \mathbf{4r^4+13r^3-13r-4} &=& \mathbf{0} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \text{Solutions: }\\ r &=& -1 & \text{no solution} \quad |r|< 1! \\ r &=& 1 & \text{no solution} \quad |r|< 1! \\ \mathbf{ r } &=& \mathbf{ \dfrac{-13+\sqrt{105}}{8} }\ \checkmark & (r=-0.34413115426) \\ \mathbf{ r } &=& \mathbf{ \dfrac{-13-\sqrt{105}}{8} }\ \checkmark & (r=-2.90586884574) \\ \hline \end{array}\)

First solution: \(\mathbf{ r =\dfrac{-13+\sqrt{105}}{8} }\)

\(\begin{array}{|rcll|} \hline \mathbf{a(1+r+r^2)} &=& \mathbf{18} \\\\ a &=& \dfrac{18}{1+r+r^2} \\\\ a &=& \dfrac{18}{ 1+ \dfrac{\sqrt{105}-13}{8}+ \left( \dfrac{\sqrt{105}-13}{8} \right)^2 } \\ \ldots \\ a &=& \dfrac{64}{13-\sqrt{105}} * \left( \dfrac{13+\sqrt{105}}{13+\sqrt{105}} \right) \\\\ \mathbf{a} &=& \mathbf{13+\sqrt{105}} & (a=23.2469507660) \\\\ x &=& a \\ \mathbf{x} &=& \mathbf{13+\sqrt{105}} \\\\ y &=& ar \\ y &=& \left(13+\sqrt{105}\right) \left(\dfrac{-13+\sqrt{105}}{8} \right) \\ y &=& -\left(13+\sqrt{105}\right) \left(\dfrac{13-\sqrt{105}}{8} \right) \\ y &=& \dfrac{-64}{8} \\ \mathbf{y} &=& \mathbf{-8} \\\\ z = ar^2 &=& yr \\ z &=& -8\left( \dfrac{-13+\sqrt{105}}{8} \right) \\ \mathbf{z} &=& \mathbf{13-\sqrt{105}} & (z=2.75304923404) \\ \\ \mathbf{x^2+y^2+z^2} &=& \mathbf{612} \\ \left(13+\sqrt{105}\right)^2 + (-8)^2 + \left(13-\sqrt{105}\right)^2 &=& 612 \ \checkmark \\ \hline \end{array} \)

Second solution: \(\mathbf{ r =\dfrac{-13-\sqrt{105}}{8} } \)

\(\begin{array}{|rcll|} \hline \mathbf{a(1+r+r^2)} &=& \mathbf{18} \\\\ a &=& \dfrac{18}{1+r+r^2} \\\\ a &=& \dfrac{18}{1+\dfrac{-13-\sqrt{105}}{8}+\left(\dfrac{-13-\sqrt{105}}{8}\right)^2} \\ \ldots \\ a &=& \dfrac{64}{13+\sqrt{105}} * \left( \dfrac{13-\sqrt{105}}{13-\sqrt{105}} \right) \\\\ \mathbf{a} &=& \mathbf{13-\sqrt{105}}& (a=2.75304923404) \\\\ x &=& a \\ \mathbf{x} &=& \mathbf{13-\sqrt{105}} \\\\ y &=& ar \\ y &=& \left(13-\sqrt{105}\right) \left(\dfrac{-13-\sqrt{105}}{8} \right) \\ y &=& -\left(13-\sqrt{105}\right) \left(\dfrac{13+\sqrt{105}}{8} \right) \\ y &=& \dfrac{-64}{8} \\ \mathbf{y} &=& \mathbf{-8} \\\\ z = ar^2 &=& yr \\ z &=& -8\left( \dfrac{-13-\sqrt{105}}{8} \right) \\ \mathbf{z} &=& \mathbf{13+\sqrt{105}} & (z=23.2469507660) \\ \\ \mathbf{x^2+y^2+z^2} &=& \mathbf{612} \\ \left(13-\sqrt{105}\right)^2 + (-8)^2 + \left(13+\sqrt{105}\right)^2 &=& 612 \ \checkmark \\ \hline \end{array}\)

Interesting question :)

\(\frac{2x+16}{x-1}=ln\left[ \frac{1}{2}\left( \frac{1}{1+sin\theta}+\frac{1}{1-sin\theta} \right) \right]\\ \frac{2x+16}{x-1}=ln\left[ \frac{1}{2}\left( \frac{1-sin\theta +1+sin\theta}{1-sin^2\theta} \right) \right]\\ \frac{2x+16}{x-1}=ln\left[ \frac{1}{2}\left( \frac{2}{cos^2\theta} \right) \right]\\ \frac{2x+16}{x-1}=ln\left[ \left(cos\theta \right)^{-2} \right]\\ \frac{2x+16}{x-1}=-2ln \left(cos\theta \right) \\ \frac{x+8}{x-1}=-ln \left(cos\theta \right) \\ \frac{x-1}{x-1}+\frac{9}{x-1}=-ln \left(cos\theta \right) \\ 1+\frac{9}{x-1}=-ln \left(cos\theta \right) \\ \frac{9}{x-1}=-ln \left(cos\theta \right) -1\\ \frac{x-1}{9}=\frac{1}{-ln \left(cos\theta \right) -1}\\ x-1=\frac{-9}{ln \left(cos\theta \right) +1}\\ x=\frac{-9}{ln \left(cos\theta \right) +1}+\frac{ln \left(cos\theta \right) +1}{ln \left(cos\theta \right) +1}\\ x=\frac{ln \left(cos\theta \right) -8}{ln \left(cos\theta \right) +1}\\\)

So it works     a=8 and b=1

Here is the question:

Logic, you have asked over 200 questions and have said thanks the grand total of one time.

I would not have worded it like guest has but I share the same sentiment.