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sqrt ( sqrt 5 a^4)

((5a^4)^1/2)^1/2 =

5^(1/2*1/2) a^(4*1/2*1/2)

5^1/4  *a

a 5^1/4

a$\sqrt[4]{5}$

Well....we could use average rates of change, but here's something that's faster

Using the function ..... 24x - 10

Setting this to 0 and solving for x we get that x  = 10/24   = 5/12

This function be negative for all x values < 5/12.....so our given function will be concave down for all the same x values

For all x values > 5/12, this function will be positive so our given function will be concave up for the same x values

So....on the interval  [ -10, -6 ]....the given  function will be concave down

Here's a graph : https://www.desmos.com/calculator/5jagkhb5zl

The problem with using rates of change to calculate concavity is that we may "skip" over some interval  where the concavity changes

Thanks Cphill

(3 √xy )^3       Write in exponential form, first

( 3^1  * x^(1/2)  * y^1)^3   =

3^3  * x^(3/2) * y^3  =

27 * x√x * y^3

11.  Notice NSS,  that taking the product of the number of volunteers with their associated time in minutes, we have that

2 * 220  =  440

5 * 88  =  440

8 * 55 +  440

11*40 + 440

So....it appears that

nt  =  440

NOW it makes a little more sense!

720 m/day  =  720 m/day x 1km/1000m = .72 km/day

1.5 gal/min =   1.5 gal/min  x  4 qt/gal =  6 qt/min   =   6 qt/min x 60min/hr = 360 qt/hr

32 yd/min = 32 yd/min x 36 in/yd x 1min/60sec = 19.2 in/sec

Can you see how it is done?   Can you do the others?

10)    Joint variation might be a new one to you, NSS

We have this

A  = kds^2      

Where A is the amount of oil used, d is the distance  and s is the speed

As in other types of variation, we need to find the proportionality constant, k

So  we know that

30  =  k (85) (42)^2         divide both sides by  (85)(42)^2

30 / [ (85)(42)^2 ]  = k    1 / 4998

So we want to find A for this

A  =  (1 / 4998) (26) (54)^2   =    15.2  barrels

They are all unit rates....for different things.

i is DEFINED as the sqrt (-1)      i x i = -1

Here is another example     sqrt (-4)  = sqrt(4 * -1)  = 2 sqrt(-1) = 2 * i  = 2i

Thank you all!!

Your answer was better than mine, killersteve.....!!!!

what is the answer for all of them 

Well CPhill Is better than me anyways. 

I guss ill stick to my welcome messages.

Good work man.

......agree   numbers that have an 'i'  (as in IMAGINARY) component ...involves a sqrt of a negative number....    SOMETIMES '0' is considered imaginary too (at least that is what I found)

This might explain it better than I can :

https://www.mathsisfun.com/numbers/imaginary-numbers.html

BTW....Isaac Asimov would argue with the term "Imaginary"....he claimed that they were just as "Real" as "Real" numbers !!!!!

Not sure what you need here.....can you add further explanation?

They are:

unit rate for speed

unit rate for volume

unit rate for speed

unit rate for speed

unit rate for speed

unit rate for volume

a number that is expressed in terms of the square root of a negative number (usually the square root of −1, represented by i or j ).

y =  1/x       y   =   5 / (x + 6)

The second graph  shifts the first graph 6 units to the left  and  vertically "stretches" it by a factor of 5

See the graphs here, NSS  :

https://www.desmos.com/calculator/ckidk8ssrf

Yah....how about  1 1/3  oz/hr  ?   ...etc   etc...etc

How about 4 cups a day

I am quite sure EP will be correct but I will take a look:

Let b and c be constants such that the quadratic $-2x^2 +bx +c$   

has roots $3+\sqrt{5}$  and $3-\sqrt{5}$.

Find the vertex of the graph of the equation   

$y=-2x^2 + bx + c$

The roots of a quadratic   $ax^2+bx+c$  is the answers to   $ax^2+bx+c=0$

And the answers to this are given by the quadratic equation

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}=3\pm\sqrt5$

This means that  $x=\frac{-b}{2a}$ = 3      must be exactly halfway between the two roots.!!

So the axis of symmetry is x=3 and the vertex lies on this line so the x value of the vertex is 3

The y value will be    $y=-2*3^2+bx+c = -18+3b+c$

This is fine but you need to find the vleu of b and c       

From the equation above I can see that 

$\frac{-b}{2a}=3 \qquad and \qquad \pm\frac{\sqrt{b^2-4ac}}{2a}=\pm\sqrt5\\ a=-2\qquad so\\ \frac{-b}{-4}=3 \qquad and \qquad \pm\frac{\sqrt{b^2+8c}}{-4}=\pm\sqrt5\\ b=12 \qquad and \qquad \pm\frac{\sqrt{b^2+8c}}{4}=\pm\sqrt5\\ b=12 \qquad and \qquad \frac{b^2+8c}{16}=5\\ b=12 \qquad and \qquad b^2+8c=80\\ b=12 \qquad and \qquad 144+8c=80\\ b=12 \qquad and \qquad 18+c=10\\ b=12 \qquad and \qquad c=-8\\$

so

$y= -18+3b+c\\ y= -18+3*12+-8\\ y=-18+36-8\\ y=10$

So the vertex is   (3,10)

Which is exactly the same answer has given you. He has done it a number of different ways but stlill all the answers are the same!!

Thanks EP :)

There are 4 basic transformations for let's say, plane figures, for the moment. They are i) translation ie sliding a figure in a particular direction. ii) reflection ie reflecting a figure about a line or if you like a mirror image. iii) rotation ie rotating or spinning a figure around a point. iv) enlargement ie changing the size of a figure by scaling up or down, leaving the figures to be similar. Except for the last transformation all the others preserve the properties of the original figure ie the image is congruent to the original figure.

One way to prove all circles are similar is take any 2 general circles as in the diagram then translating one on to the other so that the centres of each are the same point ie translate A onto B. By the definition of a circle and a translation we have 2 concentric circles. Therefore as each point on a circumference is equidistant to the centre we can say that the larger circle is an enlargement of the smaller circle by a factor of s/r for all points. 

'That's a description of how the proof might go...now it's a matter of putting it down more mathematically...which depends on what level you are at. Hope it helps you understand a bit better.

Atomic mass Cobalt ≈ 59

atomic mass Chlorine ≈ 35.5

atomic mass H₂O ≈ 18

Percentage water = 100*6*18/(59  + 2*35.5 + 6*18)

I’ll leave you to finish the calculation.