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Using distibutive property of multiplication:

=

1/2 (2x) - 1/2(10)

=

x-5

Is that the full question? If you could give the directions, and touch up the formatting it would be vey helpful!

The US seems about 65% and the UK about 15%.

Can you do it now?

You are very welcome!

:P

\(\text{If the dependence is linear, i.e. $x = \alpha y,~\alpha \in \mathbb{C}$, you can write $x \sim y$}\\ \text{Otherwise I would just write $x=f(y).\\$ I don't know of any single symbol that expresses it.}\)

A = 2( lw + lh + wh )

A  =  2lw  + 2lh  + 2wh      →    distributive property

Subtract  2lh  from both sides

A  -  2lh  =  2lw + 2wh     →   subtraction property of equality

Factor out "w"  from the right side

A - 2lh =  w [ 2l  + 2h]     

Divide both sides by  [ 2l + 2h]

[ A -2h]

_______  =  w        →      division property of eqiuality

[2l + 2h]  

Second one

f(x)  = -7x^3  - x + 1

Remember the rule .....if the lead coefficient is negative  and the power on its associated variable is odd:

The poynomial  will tend toward   + infinity   as x  →  - infinity

And....the polynomial will tend tword -infinity as x →   infinity

See the graph here : https://www.desmos.com/calculator/y47dnl4gy5

f(x)  =  2x^4 + 6x^2 + 4

The lead coefficient is positive and the power on its associated variable is even

The polynomial will approach + infinity   as x → -infinity  and as x →  infinity

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

1.What is the upper bound of the function?

f(x)=4x^4−25x^2−5x−13

I'll admit....I had this in Pre-Cal  ages ago, but I forgot how to calculate the upper and lower bounds on the zeroes without a littlte "refresher"....here are the steps to finding the upper [ and lower] bounds on the zeroes:

1. The leading coefficient must be a "1".....if not.....divide  all the terms by the leading coefficient.....so we have

x^4   - (25/4)x^2 - (5/4) - (13/4)

2.  Write down all the coefficients   →   1, (-25/4) , (-5/4) , (-13/4)

3. "Throw away" the lead coefficient , 1

4. Remove the minus signs  ....so...we are left with the values   25/4, 5/4 , 13/4

5. We now have two  possible bounds :

(a)  Bound 1  =  the largest value + 1  =  (25/4) + 1  =  (25/4) + (4/4)  =   29/4

(b)  Bound 2  = the sum of all values or 1, whichever is larger.....so...

25/4  + 5/4  + 13/4  =  43/4    which is larger than 1

The smallest of these two possible bounds is our answer  =  29/4  =  7.25

This tells us that  that the upper  and lower values on zeroes  are  -7.25   and 7.25

So....the upper bound is  7.25

Look at the graph here:  https://www.desmos.com/calculator/wdwslnqzqa

The function has two real roots.....notice that they are contained between  x = -7.25  and x  =7.25

I do not know why this works....I expect that the proof of it is difficult  !!!!

BTW :  here is a good site that reviews this : https://www.mathsisfun.com/algebra/polynomials-bounds-zeros.html

For problems like these, I like to see if we can discover a pattern

Note that  the sum of the digits from 1 -9  inclusive  = 45

And adding the "1" in the next integer, 10, gives us  46

And we can write this sum as

Sum of the individual digits from 1 - 10^1  inclusive   = 1* 45  + 1

Next....the sum of the digits of the integers from 1-99 inclusive  =  900    

We can see this because we  have the digits  1-9  in the ones place repreated 10 times  =  10(45)  = 450

And we have the sum of the digits in the tens place as  10  ( 1 + 2 +3 +....9)  = 10 (45)  = 450

And adding the "1"  in the next integer (100) gives us this sum   =  900 + 1 =    20*45 + 1  

Following this, the sum of the individual digits of the integers from 1 -999  =   13,500   [check this for yourself]

And adding the "1" in the next integer  (1000)   gives us this sum:

13500 + 1    =    300* 45  +  1

Note the pattern that seems to be emerging......the sum of the digits of the integers from 1 - 10^n  inclusive  =

(n) followed by the number of zeroes calculated as (n - 1) * 45    +  1

So....for instance .....the sum of  the digits of the integers from 1 -10^1 inclusive  =

(1)  followed by  ( 1 - 1) zeroes * 45 + 1   =

(1) followed by no zeroes * 45 + 1  =

1*45 + 1

And the sum of the digits of the integers from   1 - 10^2 inclusive  =

(2)followed by (2-1) zeroes * 45 + 1   =

(2)  followed by 1 zero  * 45 + 1  =

20*45 + 1

Note that the sum of the individual digits of the integers from 1 -100^3   =

300*45 + 1......which follows our pattern

This seems to imply that  the  sum of the individual digits in the integers from 1 - 10^100 inclusive should be :

(100) followed by (100-1) zeroes * 45 + 1  =

100 followed by 99 zeroes * 45 + 1  =

10^101 * 45 + 1  =

4.5 * 10^102  +  1       as found by the Guest and heureka  !!!!

x^2  - 5x  -  24

We  need two integers that multiply to  - 24   and sum to  - 5

- 8  and +3   will work

So....we can factor this as

(x - 8) ( x + 3)

Let A  be tthe amount that one brother has

Let  A + 15  be the amount the other brother has

And the combined amount  =   55    .....so.....

A  + ( A + 15)   =   55         simplify

2A  +  15   =    55       subtract 15 from both sides

2A  =  40       divide both sides by 2

A  = 20  

And  A + 15  =  20 +  15  =  35

So...one has saved $20  and the other has saved $35

Its ok, you seem like you need practice

You are incorrect,

that is the percentage of US drinkers

Danke  

A calculator that can display ten digits.
How many different ten-digit numbers can be type using just 0-9 keys once each,
and moving from one keypress to the next using the knight’s move in chess?
(In chess, the knight move in an L-shape – one square up and two across, two squares down and one across,
two squares up and one across, and other like combinations)

\(\begin{array}{|c|c|} \hline \text{moving from keypress} & \text{to the next} \\ \hline 0 & 3 \text{ or } 5 \\ 1 & 6 \text{ or } 8 \\ 2 & 7 \text{ or } 9 \\ 3 & 4 \text{ or } 8 \\ 4 & 3 \text{ or } 9 \\ 5 & 0 \\ 6 & 1 \text{ or } 7 \\ 7 & 2 \text{ or } 6 \\ 8 & 1 \text{ or } 3 \\ 9 & 2 \text{ or } 4 \\ \hline \end{array}\)

The graph is:

A Hamiltonian path is a path in this directed graph that visits each vertex exactly once.

Here are four Hamiltonian paths:

1.

2.

3.

4.

Hello Heureka,

yes, this is as they have it also, so I believe this is right...what I struggled with is the centre 30. If you calculated x, it comes out to be 34 and not 30. The equation for maths ends up being x-30. So saying then 34-30, we get 4, which should be math only...however we know that ALL math also take History, so what to do with the 4 then?....cast it outside?...

I do firmly understand the outcome, was just concerned about the fact that x calculated to 34 and not 30, in which case math only would have calculated to 30-30 which is zero...

thank you for helping out to understand this approach..

126 Grade 11 learners, 44 of them take Math, 112 take History and 90 take Science.

All the learners who take Math, also take History.

80 Learners take both History and Science and

30 take both Math and Science.

I assume:

du bist willkommen M elody

Just a question on this as well....if they do not give the centre value, do we accept always that that value should be calculated by using the "x" method..or should we accept that it is zero then....and omit it?...

Thank you guest, I do appreciate...

my answer got wiped ;-; I have no idea why (or am i just dum dum)

But, I think you were right. My answers came out the same as yours.

im glad to hear that you had a great day! :)) 

What is the sum total of all individual digits from 1 to 10^100? In other words:
1+2+3+4+5+6+7+8+9+[1+0]+[1+1]+[1+2]+[1+3].......and so on to 10^100?
Any help or hint would be greatly appreciated. thank you.

\(\text{Let $ \mathbf{b} = $ numeral system } \\ \text{Let $ \mathbf{s} = $ sum of all digits from $1$ to $b^n$ }\)

\(\begin{array}{|rcll|} \hline \mathbf{s} &=& \mathbf{1 + n\cdot b^n \cdot \left( \dfrac{b-1}{2} \right) } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline b &=& 10 \quad (\text{decimalism}) \\ n &=& 100 \\ && \text{from $1$ to $10^{100}$ } : \\\\ \mathbf{s} &=& \mathbf{1 + n\cdot b^n \cdot \left( \dfrac{b-1}{2} \right) } \\ s &=& 1 + 100\cdot 10^{100} \cdot \left( \dfrac{10-1}{2} \right) \\ &=& 1 + 10^2\cdot 10^{100} \cdot \left( \dfrac{9}{2} \right) \\ &=& 1 + \left( \dfrac{9}{2} \right)\cdot 10^{102} \\ \mathbf{s} &=& \mathbf{ 1 + 4.5 \cdot 10^{102} } \\ \hline \end{array}\)