All Answers 358512 Answers

2/3 t -1

-1 <1/3 t +7

-8 < 1/3 t

-24 < t

t+7 <= -2t+15

3t <= 8

t <= 8/3

(-24,8/3]

I don't believe we can assume that a = b = c, since all we are given is that their three moduli are equal and nonzero. So I will provide an answer that does not reqire that assumption.

We are given that \(az^2+bz + c = 0\), so we can assume

\(0= |az^2+bz+c| \geq|az^2+bz|-|c| \geq|az^2|-|bz|-|c|\).

But then

\(|a||z|^2-|b||z|-|c|=k|z|^2-k|z|-k=k(|z|^2-|z|-1)\leq0\),

where \(k=|a|=|b|=|c|\)is nonzero and positive.

This forces the inequality \(|z|^2-|z|-1\leq0\).

From the graph of the real-valued function \(f(|z|)=|z|^2-|z|-1\) given above, one can see that the values of \(|z|\) for the portion of the graph below and touching the \(|z|\) axis range from

|z|= \(\frac{1-\sqrt{5}}{2} \) to |z|=\(\frac{1+\sqrt{5}}{2} \). So the largest value of |z| is \(\frac{1+\sqrt{5}}{2} \).

You post the same thing, If you plan on posting at least post something useful

melody doesnt need your daily remainder, she's smart enough to know when to troll the lazy brats with troll answers

I think that you should make an attempt at solving it and then show us what you have done, and then we will check out your work and see if you got it right.

There maybe many solutions to your probelem. You didn't state the number of digits in each integer.

For example, here are 4 consecutive 5-digit integers that satisfy all the congruences:

12130 mod 5=0, 12131 mod 7=0, 12132 mod 9=0, 12133 mod 11=0 

12130 + 12131 + 12132 + 12133 = 48,526

P.S. Here are 4 consecutive 4-digit integers that also satisfy all the congruences. According to computer search it seems to indicate these are the smallest 4 consecutive integers:

1735 mod 5 =0,  1736 mod 7=0,  1737 mod 9 =0,  1738 mod 11 =0

1735 + 1736 + 1737 + 1738 = 6,946.

(1735, 5200, 8665, 12130, 15595, 19060, 22525, 25990, 29455, 32920, 36385, 39850, 43315, 46780, 50245, 53710, 57175, 60640, 64105, 67570, 71035, 74500, 77965, 81430, 84895, 88360, 91825, 95290, 98755)>>Total = 29 numbers up to 100,000 that satisfy the 4 congruences. However, there are 289 numbers up to 1,000,000 that satisfy the 4 congruences. Each number begins with mod 5=0. Then 1 is added to the follwing 3 numbers. For example:5200 mod 5, 5201 mod 7, 5202 mod 9 and 5203 mod 11 =0.

sorry, I was doing it and I am almost there.

I see you have so far never responded to any of your answerers. This is definitely not a good look!

One gets a bite, the other gets the rest.

Here is a good hint, and I do not think you should be given more.

It is this type of puzzle:

This is not yours, I just found a similar puzzle pic on line. But  I would tackle the problem with a grid like this one.

I have not done it, I am just assuming that this would be the right approach.

You put a cross in all the ones that do not work and a tick in the ones that do work until you have averything filled in.

You start by putting a tick in the red house box, which means that you can cross all the other colour houses. etc.

Your teacher probably already told you this much anyway.

Good luck.

Come on Jenny.

This is a challenge question that your teacher has given the class.

Maybe there is some prize, or at least prestige, for getting it nutted out.

Do you really think it is right for you to get other people to do it and present it as your own?

Over 23 housr later and no response from SinclairDragon.428

Why does this not surprise me.

Thanks Jenny  :)

Using the equation  of the line

2x + 3y  = 3

3y = 3  - 2x

y =  (3 - 2x)  /3

Let  the center of the circle be  ( x, (3 -2x) / 3  )

And  the distance from this center  to both of the given  points is  equal ....so...

(x -8)^2  + [ (3 -2x)/3  + 5]^2   =   ( x + 1)^2 + [ (3 -2x)/3 -4 ] ^2    simplify

x^2 - 16x + 64  +  [ (18 -2x)/ 3 ]^2  =  x^2 + 2x + 1 +  [ (-9 - 2x) / 3 ] ^2

x^2 - 16x+ 64 + [ 324 - 72x + 4x^2/] /9  = x^2 + 2x + 1  + [ 81 + 36x + 4x^2 ] / 9

-16x + 64  + [ 324 - 72x + 4x^2 ] / 9  =  2x + 1 + [81  +36x+ 4x^2]/9      multiply through by 9

-144x + 576 + 324 - 72x + 4x^2 =  18x + 9 + 81 + 36x + 4x^2

-216x + 900 = 90 + 54x

810  = 270 x     divide both sides by 270

3  = x

And using the equation of the line     y = [ 3 - 2(3)] / 3  = -1

So...the center of the circle is  (3 , -1)

And using either point and this center, we can find r^2  as

(8 - 3)^2 + ( -5 + 1)^2  = r^2

5^2 + (-4)^2 = r^2

41  = r^2

So...the eqaution is  ( x - 3)^2 + ( y + 1)^2  =  41

Here's the graph : https://www.desmos.com/calculator/20kbuxcqtc

No  parentheses  around the "1"

g(x) = log ₅  ( x - 1)  is a shift of the parnet unit to the right by one unit

g(x)  =log 5 (x)   - 1   is a shift of  the parent function down one unit

We want the second one.....

for the first 1 theres two options that are very similar, would you need the parenetheses around (x+1) or no?

OM = 2cm and O'M = 2cm so OO' = 2sqrt(2) cm.  O'P = 2cm, so radius of large circle = OO' + O'P = 2(1+sqrt(2))cm.

thanks dear 

The midpoint  of the two points will be the center of the  circle

So

( [ 2 + 6 ] / 2 , [ -3 + 5 ] /2 )   =   ( 4 , 1)   is the center

To find the radius^2   use the center and either point...so we have

(4 - 6)^2 + ( 1 - 5)^2  = r^2

(2)^2 + (4)^2  = r^2

4 + 16  = r^2

20  = r^2

So....the equation is

(x - 4)^2 + ( y - 1)^2  = 20

See the graph here : https://www.desmos.com/calculator/xl9ckproc3

I get a = 3 and b = 4.

Let's write

y = 5^(x + 1)           we wnat to get x by itself and then "swap"  x and y

In log form we have

log ₅ (y)  = x + 1       subtract 1 form  both sides

log ₅ (y) - 1  = x     "swap" x and y

log ₅ (x)  - 1  = y       this is the inverse   and we can write

g(x)  = log ₅ (x)  - 1    [ this goes in the first blank ]

The graph of the exponential function is a shift of  1 unit   to the left

The graph of the log function  is a shift  1 unit down from its parent function

I don't know if you mean 1.4 times -3 times 5, or if this is question 1 as 4*-3*5.

If it was 1.4, then...

-21 is the answer

And if not, then multiply 4*3*5 and add a negative.

You are very welcome!

:P

sqrt [ 17 + 12sqrt (2)  ] 

Let us suppose that we can write this in the form a + b sqrt (2)    where a, b are both positive integers

 So

sqrt [ 17 + 12sqrt(2) ]  = a + b sqrt (2)       square both sides

17 + 12 sqrt (2)  =  a^2 + 2ab sqrt (2)  + 2b^2

Equating coefficients it must be that

12sqrt (2)  =  2ab sqrt (2)         and     a^2 + 2b^2  = 17      (2)

12  = 2ab

6  = ab

b = 6/a      (1)

Sub (1)  into (2)  and we have that

a^2 + 2 (6/a)^2 =17

a^2 + 2*36 / a^2  = 17

a^2 + 72/a^2  = 17       multiply through by a^2

a^4 + 72  = 17a^2

a^4 - 17a^2 + 72  =  0        factor

(a^2 - 8) ( a^2 - 9)  =  0

We want  a  to be a positive integer  so  a =  3

And b = 6/a=  6/3  =2

So.....sqrt [ 17 + 12sqrt (2)  ]  can be simplified to

3 + 2sqrt (2)