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what is 0.0000686 in scientific notation

It is a very little number so the index will be negative.

You want just one non zero digit in front of the point so the number part will be 6.86

You had to jump the point over 4 zeros and the 6 to get it to its new position.  That is 5 jumps altogether.  So the power will be 5 BUT remember it is negative!

\(0.0000686=6.86\times 10^{-5}\)

= 5x + (5*3)/5 -3

= 5x + 15/3 -3

= 5x + 5 - 3 = 5x + 2

Each ticket costs  $5214/66 tickets

So SEVEN tickets would cost:

5214/66   x  7  = $ 553

Wise Old Owl says "Three"

Thanks heaps!!

Greatly appreciated!

:)

x+9 over x-4 equals 7 over 9 What does x equal ?

\(\begin{array}{rcll} \frac{x+9}{x-4} &=& \frac79 \qquad & &| \qquad\cdot 9\\\\ 9\cdot \left( \frac{x+9}{x-4} \right) &=& 7 \qquad & &| \qquad\cdot (x-4)\\\\ 9\cdot(x+9) &=& 7\cdot (x-4) \\\\ 9x + 9\cdot 9 &=& 7x - 7\cdot 4 \\ 9x + 81 &=& 7x - 28 \qquad & &| \qquad -7x \\ 9x -7x + 81 &=& - 28 \\ 2x + 81 &=& - 28 \qquad & &| \qquad -81 \\ 2x &=& - 28 -81 \\ 2x &=& -109 \qquad & &| \qquad :2 \\ x &=& -\frac{109}{ 2 }\\ \mathbf{x} &\mathbf{=}& \mathbf{-54.5}\\ \end{array}\)

Solve for x, showing your work every step of the way.

3(x + 4) + 5x - 7 = 6 + 2x - 2(x - 4)

\(\begin{array}{rcll} 3(x + 4) + 5x - 7 &=& 6 + 2x - 2(x - 4) \\ 3x + 3\cdot 4 + 5x - 7 &=& 6 + 2x - 2x -2\cdot (-4) \qquad & | \qquad 3\cdot 4 = 12\\ 3x +12 + 5x - 7 &=& 6 + 2x - 2x -2\cdot (-4) \qquad & | \qquad -2\cdot (-4) = +8\\ 3x +12 + 5x - 7 &=& 6 + 2x - 2x +8 \qquad & | \qquad 2x - 2x = 0\\ 3x + 12 + 5x - 7 &=& 6 + 0 +8\\ 3x + 12 + 5x - 7 &=& 6 +8 \qquad & | \qquad 6 +8 = 14\\ 3x + 12 + 5x - 7 &=& 14 \\ 3x + 5x + 12 - 7 &=& 14 \qquad & | \qquad 12 - 7= 5\\ 3x + 5x + 5 &=& 14 \qquad & | \qquad 3x + 5x= 8x\\ 8x + 5 &=& 14 \qquad & | \qquad -5\\ 8x &=& 14 -5 \qquad & | \qquad 14 -5= 9\\ 8x &=& 9 \qquad & | \qquad :8\\ x &=& \frac{9}{ 8 }\\ \mathbf{x} &\mathbf{=}& \mathbf{1.125} \end{array}\)

85.6 as a mixed number

\(85.6 = 85 + \frac{6}{10}\\ = 85 + \frac{3}{5}\\ = 85 \frac{3}{5}\\\)

a rectangle has a width of 45 centimeters and a perimeter of 188 centimeters. What is the rectangles length?

perimeter p = 188 cm

width w = 45 cm

length l = ? cm

\(\begin{array}{rcll} p &=& 2w + 2l \qquad & | \qquad w = 45\qquad p= 188\\ 188 &=& 2 \cdot 45 + 2l \\ 188 &=& 90 + 2l \qquad &| \qquad -90\\ 188-90 &=& 2l \\ 98 &=& 2l \\ 2l &=& 98 \qquad &| \qquad :2\\ l &=& \frac{98}{ 2} \\ l &=& 49 \\ \mathbf{\text{length} }&\mathbf{=}& \mathbf{49\ \text{cm}} \end{array}\)

slope of 9,3 19,-17

\(\begin{array}{rcll} \text{slope } &=& \frac{ \Delta y } { \Delta x } \\ &=& \frac{-17-3}{19-9} \\ &=& \frac{-20}{10} \\ \mathbf{ \text{slope } } &\mathbf{=}& \mathbf{-2} \\ \end{array}\)

\(\frac{9.45\times 10^{10}}{1.5*10^6}\\ =\frac{9.45}{1.5}\times \frac{10^{10}}{10^6}\\ =\frac{18.9}{3}\times \frac{10^{4}}{1}\\ =6.3\times 10^4\\ or\\ =63,000\)

-5x + 2 – 4x = 47 Solve for X.

\(\begin{array}{rcll} -5x + 2 – 4x &=& 47 \\ -5x – 4x + 2 &=& 47 \\ -9x + 2 &=& 47 \qquad &| \qquad \cdot(-1)\\ 9x - 2 &=& -47 \qquad &| \qquad +2\\ 9x &=& -47 +2\\ 9x &=& -45 \qquad &| \qquad :9\\ x &=& -\frac{45}{9} \\ x &=& -\frac{5\cdot 9}{9} \\ \mathbf{x } &\mathbf{=}& \mathbf{-5} \end{array}\)

How to proof a+a*2=?

\(a+a\cdot 2 = a\cdot(1+2) = a \cdot 3 = 3a\)

Factor the following expression:

5x^2+10x-840

\(\begin{array}{lrcl} &5x^2+10x-840\\ &= 5(x^2+2x-168) \\ \text{Factorisation }& =5(x-12)(x+14)\\ &\mathbf{x_1 = 12}\\ &\mathbf{x_2 = -14} \end{array}\)

Determine where f(x) intersects the x-axis ( f(x) = 0 ).

f(x) = 2x^2+10x+12

\(\begin{array}{rcl} &2x^2+10x+12 &=& 0\\ &2(x^2+5x+6) &=& 0\\ \text{Factorisation }& 2(x+2)(x+3) &=& 0\\ &x_1 &=& -2\\ &x_2 &=& -3 \end{array}\)

A group of adults and kids went to see a movie. Tickets cost $7.00 each for adults and $4.50 each for kids, and the group paid $64.00 in total. There were 4 fewer adults than kids in the group.
Find the number of adults and kids in the group.

\(\begin{array}{lrcl} a = \text{adults}\\ k = \text{kids}\\ \hline (1) & a\cdot $7.00 + k\cdot $4.50 &=& $64.00\\ (2) & a &=& k-4 \\ \hline & a\cdot $7.00 + k\cdot $4.50 &=& $64.00 \qquad | \qquad a &=& k-4\\ & (k-4)\cdot $7.00 + k\cdot $4.50 &=& $64.00\\ & k\cdot $7.00 -4\cdot $7.00 + k\cdot $4.50 &=& $64.00\\ & k\cdot $7.00 + k\cdot $4.50 -4\cdot $7.00&=& $64.00\\ & k\cdot $11.5 -4\cdot $7.00&=& $64.00 \\ & k\cdot $11.5 - $28.00&=& $64.00 \qquad | \qquad + $28.00\\ & k\cdot $11.5 &=& $64.00 + $28.00\\ & k\cdot $11.5 &=& $92.00 \qquad | \qquad : $11.5\\ & k &=& \frac{ $92.00}{ $11.5 } \\ & \mathbf{k} &\mathbf{=}& \mathbf{8} \\ \hline & a &=& k-4 \qquad | \qquad k = 8\\ & a &=& 8-4 \\ & \mathbf{a} &\mathbf{=}& \mathbf{4} \\ \end{array}\)

The number of adults is 4

and the number of kids in the group is 8

what is the unit digit in 2^2015

\(\begin{array}{rcll} 2^{2015} \pmod {10}\\ &\equiv& 2^{13\cdot 155} \pmod {10} \qquad \boxed{~2^{13} \pmod {10} \equiv 8192 \pmod {10}\equiv 2 \pmod {10}~} \\ &\equiv& \left(2^{13}\right)^{155} \pmod {10} \qquad | \qquad 2^{13} \pmod {10}\equiv 2 \pmod {10}\\ &\equiv& 2^{155} \pmod {10}\\ &\equiv& 2^{13\cdot 11+12} \pmod {10}\\ &\equiv& 2^{13\cdot 11}\cdot 2^{12} \pmod {10}\\ &\equiv& \left(2^{13}\right)^{11}\cdot 2^{12} \pmod {10} \qquad | \qquad 2^{13} \pmod {10}\equiv 2 \pmod {10}\\ &\equiv& 2^{11}\cdot 2^{12} \pmod {10}\\ &\equiv& 2^{11+12} \pmod {10}\\ &\equiv& 2^{23} \pmod {10}\\ &\equiv& 2^{13+10} \pmod {10}\\ &\equiv& 2^{13}\cdot 2^{10} \pmod {10} \qquad | \qquad 2^{13} \pmod {10}\equiv 2 \pmod {10}\\ &\equiv& 2\cdot 2^{10} \pmod {10}\\ &\equiv& 2^{11} \pmod {10} \\ &\equiv& 2048\pmod {10}\\ &\equiv& 8\pmod {10}\\ \mathbf{2^{2015} \pmod {10}} & \mathbf{\equiv} & \mathbf{8\pmod {10}}\\ \end{array}\)

4:5=6:x ?

\(\begin{array}{rcll} 4:5 &=& 6:x \qquad \text{or}\\ \frac45 &=& \frac6x \qquad \updownarrow\\ \frac54 &=& \frac{x}6 \qquad | \qquad \cdot 6\\ 6\cdot \frac54 &=& x \\ x &=& 6\cdot \frac54 \\ x &=& 3\cdot \frac52 \\ x &=& \frac{15}2 \\ \mathbf{x} &\mathbf{=}& \mathbf{7.5} \\ \end{array}\)

Hi Omi67 :)

Sorry!

Maybe you are missing brackets? The answer is h=0

The easiest way to start it is to multiply both sides by 5.

This gives you 10h+1=20h+1

 3^20=9^10=81^5

Now if you multiply 2 numbers together that both and in 1 then the last digit must be 1.   

That is what they wanted you to say :)

I am not CPhill

But Hi.  :)))

|2x+5| = 3x+4

\(\begin{array}{rcl} |2x+5| &=& 3x+4 \qquad | \qquad ()^2\\ (2x+5)^2 &=& (3x+4)^2 \\ 4x^2+20x+25 &=& 9x^2+24x+16\\ 5x^2+4x-9 &=& 0 \\\\ \boxed{~ x = {-b \pm \sqrt{b^2-4ac} \over 2a} ~}\\ 5x^2+4x-9 &=& 0 \\ x_{1,2} &=& {-4 \pm \sqrt{16-4\cdot 5 \cdot(-9)} \over 2\cdot 5}\\ x_{1,2} &=& {-4 \pm \sqrt{16+180} \over 10}\\ x_{1,2} &=& {-4 \pm 14 \over 10}\\\\ x_1 &=& {-4 +14 \over 10}\\ x_1 &=& {10 \over 10}\\ \mathbf{x_1} &\mathbf{=}& \mathbf{1}\\ x_2 &=& {-4 -14 \over 10}\\ x_2 &=& {-18 \over 10}\\ \mathbf{x_2} &\mathbf{=}& \mathbf{-1.8} \qquad \text{no solution}\\ \end{array}\)