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CPhill has already answered this at https://web2.0calc.com/questions/hey-new-topic-i-need-help-with

101^99    =   (100   + 1)^99

Binomial expansion

C(99,0)*100^99   +  ...........+  C (99.98) (100)   +  C(99,99) *1

The  last two  terms  will give us the  last  3 digits

C(99,98) * 100  +  C(99,99) * 1  =

9900  +  1  =

901

I don't think those equations are possible.

=^._.^=

If \(x <5\), then \(\left |x-5 \right |<0\). \(\left |x-5 \right |\) would be \(5-x\). So \(8x+5-\left |x-5 \right |\) would be \(9x\).

What is the slope of any line perpendicular to 6x+3y-1=0?

Hello Guest!

\(6x+3y-1=0\\ y=-\frac{6}{3}x+\frac{1}{3}\\ y=-2x+\frac{1}{3}\)

           m        n

\(m_{perpendicular}=-\frac{1}{m}=-\frac{1}{-2}=\color{blue}\frac{1}{2}\)

  !

6x+3y-1=0

3y = -6x + 1

y = -2x + 1/3

The slope is -2, so the perpendicular slope line is the negative reciprocal, 1/2. 

=^._.^=

f(1) = 4

f(1+1) = f(1) + f(1) + 1*1

f(2) = 9

f(2+2) = f(2) + f(2) + 2*2

f(4) = 22

f(4+4) = f(4) + f(4) + 4*4

f(8) = 60

=^._.^=

what does X equal? -5(X-31)=11.5

Hello niyirixu!

\(-5(x-31)=11.5\ |\ multiply\ by\ the\ terms\ of\ the\ bracket \)

\(-5x+155=11.5\ |\ -155\ on\ either\ side\ of\ the\ equation\)

\(-5x=11.5-155\\ -5x=-143.5\ |\ :(-5)\ |\ on\ either\ side\ of\ the\ equation\\ x=-143.5:(-5) \)

\(x=28.7\)

  !

-5(X-31)=11.5

x - 31 = -2.3

x = 28.7

=^._.^=

A number is a palindrome if it reads the same way forwards and backwards.
For example, 27872 and 48484 are palindromes, but 28389 and 12541 are not.
What is the sum of all of the 5-digit palindromes?

I assume:

\(\begin{array}{|l|r|r|} \hline & & \text{sum of $5$-digit palindromes} \\ \hline \text{sum until $1$-digit palindromes} & 45 & \\ \hline \text{sum until $2$-digit palindromes} & 540 & \\ \hline \text{sum until $3$-digit palindromes} & 50040 & \\ \hline \text{sum until $4$-digit palindromes} & 545040 & \\ \hline \text{sum until $5$-digit palindromes} & 50045040 & \mathbf{49500000} \\ \hline \end{array}\)

10+12+9+4=35

1.  Divide both sides by 1000000 to get 211 = 0.9^t

2.  Take logs of both sides:   ln(211) = ln(0.9^t)        where ln is log to the base e (though you could use log to the base 10 if you prefer).

3.  Use a property of logs:  ln(211) = t*ln(0.9)

4.  Rearrange:   t = ln(211)/ln(0.9)

I'll leave you to evaluate (you can use the calculator on the Home page here).

Solution:

If the arc of LM is 72° on the larger circle, then the arc LPM on the larger circle is

360 - 72 = 288°

Am I missing something?   

--. .-

\({x+2 \over x^2-x-2} = {x+2 \over (x-2)(x+1)} = {A\over x-2}+{B\over x+1}\)

Multiplying by (x - 2)(x + 1) both sides, 

\(A(x+1)+B(x-2)=x+2\)

Put x = -1  ⇒ -3B = 1

                      B = -1/3

Put x = 2  ⇒  3A = 4

                     A = 4/3 

Thus the answer is \(({4\over 3},{-1\over 3})\). 

Thanks Bginner, that is a pretty neat way to look at it.

What Bginner means is if you write them all in base2 and then add all of them together in base 2 then you will have the biggest possible number of digit combinations.

I am not explaining it very well and I don't want to give the answer that i got.  So think about it.

Let the set of values of r = {4,5} and k = {2,3,4,5,6,7} and let E be the event that  \(r\over k\)  is an integer. 

⇒Total no. of outcomes = 12 

No. of favourable outcomes = 2 

∴ \(P(E) = {2\over 12} = {1\over 6}\)

Digits are 1, 2, 3, 4 and 5. 

⇒no. of 1-digit odd no.'s  = 3

           [Repetition of digits not allowed] 

No. of 2-digit odd numbers

          one's place has = 3 possibilities 

          Then, 10's place has = 4 possibilities

⇒ no. of 2-digit odd numbers = 4 × 3  = 12 

No. of 3-digit odd numbers

           one's place has = 3 possibilities 

           10's place has = 4 possibilities 

           then, 100's place has = 5 - 2 = 3 possibilities 

 ⇒ no. of 3-digit odd numbers = 3 × 4 × 3 = 36 

No. of 4-digit odd numbers

           one's place has = 3 possibilities

           10's place has = 4 possibilities

           100's place has = 3 possibilities

           then, 1000's place has = 2 possibilities

⇒ no. of 4-digit odd numbers = 3 × 4 × 3 × 2 = 72

No. of 5-digit odd numbers

           one's place has = 3 possibilities

           10's place has = 4 possibilities

           100's place has = 3 possibilities

           1000's place has = 2 possibilities

           then, 10000's place has = 1 possibility

⇒ no. of 5-digit odd numbers = 3 × 4 × 3 × 2 × 1 = 72

∴ No. of odd numbers = 72 + 72 + 36 + 12 + 3 = 195

ABCDEF is a regular hexagon and PQR is an equilateral triangle.

In ABCDEF, 

area(△AOB) = area(△BOC) = area(△COD) = area(△DOE) = area(△EOF) = area(△FOA) 

⇒area(ABCDEF) = area(△AOB + △BOC + △COD + △DOE + △EOF + △FOA) 

                            \(= {\sqrt3 \over 4}+ {\sqrt3 \over 4}+ {\sqrt3 \over 4}+ {\sqrt3 \over 4}+ {\sqrt3 \over 4}+ {\sqrt3 \over 4}\)

                            \(=6× {\sqrt3 \over 4}\)

                            \(={3\sqrt3 \over 2}\) sq. units

⇒area(△PQR) \(={\sqrt3 \over 4}×4\)

                        \(=\sqrt3\) sq. units

\({area(ABCDEF)\over area(△PQR)} = {3\over 2} = 3:2\)

 f (x) =   -5x³   + 10x² .

The x-intercept of  the function is at x = 2

Let n(F) choose football, n(T) in tennis, n(R) in running, n(B) boys and n(G) girls

• n(F∪T∪R) = 115
• n(B) = 54    ⇒n(G) = 115 - 54 = 61
• n(F) = 52    ⇒n(F∩G) = 25    so    n(F∩B) = 52 - 25 = 27
• n(T) = 43
• n(R∩G) = 10

No. of girls in tennis

              n(T∩G) = n(G) - [n(F∩G) + n(R∩G)] = 61 - (25 + 10) = 26

No. of boys in tennis

              n(T∩B) = n(T) - n(T∩G) =  43 - 26 = 17 

∴ Remaining no. of boys in running

              n(R∩B) = n(B) - [n(F∩B) - n(T∩B)] = 54 - (27 + 17) = 10

⇒n(R) = n(R∩G) + n(R∩B) = 10 + 10 = 20

Probability of choosing running 

            \(P(R) = {20\over 115} = {4\over 23}=0.17\)

Idk but how to solve unless the radius is given? Circle LOM is irrelevant to the question. 

What's the question?

\(x+y=10\)

\(x^2+y^2+2xy=100\)

\(2xy=100-52\)

\(2xy = 48\)

\(xy=24\)

⇒\(x+{24\over x}=10\)

\(x^2-10x+24=0\)

\(x=6,4\)

1st Case : when x = 6 ⇒y = 4 

2nd Case : when x = 4 ⇒y = 6 

∴ All pairs of solutions are (6,4) and (4,6) 

You should think in terms of base-2 representation of nonnegative integers.

\begin{eqnarray*}
    f(2) &=& 1.33333 \\
    f(2)\cdot f(3) &=& 1.71429 \\
    f(2)\cdot f(3) \cdot f(4) &=& 2.10989
\end{eqnarray*}
This table shows that the largest such $n$ is 3.