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A function f is even if the graph of f is symmetric with respect to the y-axis. Algebraically, f is even if and only if f(-x) = f(x) for all x in the domain of f. Afunction f is odd if the graph of f is symmetric with respect to the origin.

no -7 didn't work can you tell me how to complete it please?

So the answer would be -7?

Same process as for https://web2.0calc.com/questions/the-quadratic-x-2-20x-36-can-be-written-in-the-form, except you now compare the coefficient of the x^2 term as well.

a(x - h)^2 + k → a(x^2 - 2hx + h^2) + k → ax^2 - 2ahx + ah^2 +k

coeff of x^2:   a = 3

coeff of x:      -2ah = -6

constant;      ah^2 + k = -2

I'll leave you to complete it!

"The quadratic x^2-20x+36 can be written in the form (x+b)^2+c where b and c are constants.  What is b+c"

Expand (x+b)^2 + c → x^2 + 2bx + b^2 + c

Compare with x^2 - 20x + 36

Coefficient of x term:    2b = -20  hence b = -10

Constant term:    b^2 + c = 36,    (-10)^2 + c = 36,    100 + c = 36,   c = -64

b + c = -74

Here is another couple of possibilities :)

OR

49229 = 19*2591

Hi Will, it is good to see you on the forum again :)

5 * (5x + 1)

= 5 * 5x + 5 * 1

= 25x + 5

Yes, you got it right.

The main concern we have in solving this problem is to find the interest rate. But, since you have 4 out of 5 variable, that is: FV=$1,000,000, PV=$25,000, Periodic Payments=$7,500 each, and N=30 Number of periods, then its relatively simple to find the interest rate using iteration. We could try a rate that will equate the FV of $25,000 Plus the FV of the stream of payments to =$1,000,000.

But modern computers and calculators can find such rate almost instantaneously once we enter all the variable we have accurately. For that purpose, I shall use this fine online calculator to do that:

https://arachnoid.com/finance/

We simply enter $1,000,000 under FV, $25,000 as (-) under PV, $7,500 also as (-) under PMT and 30 under NP. The two negative ones are "Payments made or owed". Then we simply press "IR", or Interest rate and it instantly gives us the Interest Rate we are after. Keep in mind that we have to use the "beginning" instead of the "End" of the periods since they are made at the beginning of the year.

Once we enter them and press "IR", it instantly gives us a rate of:7.254922%

And now it becomes a simple matter of finding his "Perpetual Payment", by simply multiplying this by his $1,000,000 as follows: $1,000,000 x $7.254922% =$72,549.22 - his annual perpetual payment.

P.S. Learn to use the above online calculator to solve virtually all the TVM problems that you may have.

What you got is the number's square root. Its two "multiples", by which you mean its "factors" are as follows:

1508629006561 = 53 * 28464698237

Simplify 36:72 in ratio 

Divide both by 36

1:2

1228262.59674 * 1228262.59674 = 1508629006561

Figured it out.

See https://web2.0calc.com/questions/i-need-help-on-graphing-piecewise-limits-with-infinity#r1 

.

I should have expanded my answer a tad bit more

We will have a vertical asymptote  ar x  =  -8

As the graph approaches x = -8 from the left , the graph will approach infinity

As the graph approaches x = -8 from the right , the graph will approach  -infinity

And there is a horizontal asymptote at y = 3  which is not crossed by the graph

So.....the range is   ( - infinity, 3) U ( 3 , infinity )

B           C              A            B            B         A           C        B          B       C

       1            →            2            →          3          →         4        →        1

A           D              D            C           C         D           D        A          A       D

Note that  every 4  "transformations"  gets us back to the original orientation

So.......

Dividing  2007 by 4...

We get a remainder of  3/4  which results in the 3rd orientation in the pattern, i.e.,

CBAD

√[ 4x - 3 ]  + 10 / √[ 4x - 3]  = 7

Let √ [ 4x - 3]  = a

So we have

a + 10/a  = 7           multiply through by a

a^2 + 10  = 7a      subtract 7a from both sides

a^2 - 7a + 10  = 0     factor

( a - 5) (a - 2)  = 0

Setting each factor to 0 and solving produces a = 5  or a  = 2

So..... 

√ [ 4x - 3]  =  5      square both sides            

4x - 3  = 25           add 3 to both sides

4x  = 28                divide both sides by 2

x  = 7

or

√ [ 4x - 3] = 2   square bot sides

4x - 3  =   4      add 3 to both sides

4x  = 7           divide both sides by 4

x =  7/4

Hectictar's answer is correct.....I mis-read  the problem....!!!

Here's another interpretation.

a + 1 + 4 + c  =  28       →       a + c  =  23

a + 3 + 5 + d  =  28       →       a + d  =  20

b + 4 + 2 + d  =  28       →       b + d  =  22

b + 1 + 3 + e  =  28       →       b + e  =  24

c + 2 + 5 + e  =  28       →       c + e  =  21      Add all five of these equations together to get...

2a + 2b + 2c + 2d + 2e  =  110      Divide through by  2 .

a + b + c + d + e  =  55

What is the range of the function (3x+1)/(x+8)

I am not sure how CPhill looked at this question. This is what I'd do.

Chris may well have an easier way to look at it  

x cannot equal -8 because the denominator cannot be equal to zero.

If I do the algebraic division here then   

$f(x)=\frac{3x+1}{x+8}=3+\frac{-23}{x+8}\\ Now\;\;\frac{-23}{x+8}\;\;\text{cannot equal 0 so } f(x)\ne 3\\ Also\\ y=\frac{-23}{x+8}+3\\ \text{is of the from }y=\frac{a}{x-h}+k \\ \text{so it must be a hyperbola with asymptotes at x=-8 and y=3.}$

the range Chris gave is certainly correct, I just wanted to discuss the question more. :)

1) Consider the equation 2u + v + w + x + y + z = 3. How many solutions (u , v , w , x , y , z) of non-integers does this equation have?

It has an infinite number of solutions

What is the domain of the function f(x) = sqrt6x-3 

Do you mean

$f(x) = \sqrt6*x-3 \qquad \text {Which is what you asked}\\ domain \;\;x\in R\\~\\ or\\~\\ f(x) = \sqrt{6x}-3 \\ domain\;\;x\ge 0\qquad x\in R$

If "t" has to be an integer...it is impossible to have integer sides with "4" as a hypotenuse

So.......I will assme that either

4^2 + ( t - 1)^2  = ( t + 1)^2     or    4^2 + (t + 1)^2  = (t - 1)^2

But.......for the second to be true, t  = -4, which would give us negative side lengths

So.....we have

4^2 + ( t - 1)^2  = ( t + 1)^2

16 -2t  =  2t

16  = 4t   →   t  = 4

So....the other two sides are 3 and 5

If there are no restrictions on t, we have that

( t - 1)^2  +  ( t + 1)^2  = 4^2

2t^2  + 2  =  16

t^2 =  7   →   t =  √7

So.....the sides are   √7 - 1 , √7 + 1 ,  4