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How i find out Anti log3.237363

x + 2y = 4

2x - y = 13

\(\small{ \begin{array}{lrcl} (1) & x+2y &=& 4 \\ & x &=& 4 - 2y \\\\ (2) & 2x-y &=& 13 \\ & 2(4-2y) -y &=& 13 \\ & 8 -4y - y &=& 13 \\ & 8 - 5y &=& 13 \qquad | \qquad +5y-13\\ & 8-13 &=& 5y \\ & -5 &=& 5y \qquad | \qquad :5\\ & -1 & =& y\\ & \mathbf{y} &\mathbf{=}& \mathbf{-1}\\\\ & x &=& 4 - 2(-1) \\ & x &=& 4 + 2\\ & \mathbf{x} &\mathbf{=}& \mathbf{6} \end{array} }\)

Line AB has the equation 3x - 4y + 5 = 0. Point (p, p+2) lies on line AB. What is the value of p?

\(\small{ \begin{array}{rcl} 3x-4y+5 &=& 0\\ \\ \hline \\ x &=& p \\ y &=& p+2 \\ \\ \hline \\ 3(p)-4(p+2) + 5 &=& 0\\ 3p - 4p -8 +5 &=& 0\\ -p -3 &=& 0\\ p + 3 &=& 0\\ \mathbf{p} &\mathbf{=}& \mathbf{- 3} \end{array} }\)

The value of p is -3.

The Point (-3,-1) lies on line AB

Which equation would you find to solve 9ˣ=15 for x?

\(\begin{array}{rcll} 9^x&=&15\\ 3^{2x} &=& 3\cdot 5 \qquad & | \qquad \ln{()}\\ 2x\ln{(3)} &=& \ln{(3)} + \ln{(5)} \qquad & | \qquad : \ln{(3)}\\ 2x &=& 1 + \frac{ \ln{(5)}}{ \ln{(3)}}\\ x &=& \frac12 \left[ 1 + \frac{ \ln{(5)}}{ \ln{(3)}} \right]\\ \mathbf{x} & \mathbf{=} & \mathbf{1.23248676036} \end{array} \)

Hallo guest

Hallo guest !

i think is 3

Hallo guest !

Probability that the first sweet is orange = 6/n

Probability that, if the first was orange, the second is also orange = 5/(n-1)  beacuse there are only 5 orange sweets left and there are only n-1 sweets left in total.

Hence overall probability that her two sweets are orange = (6/n)*(5/(n-1)) = 30/(n(n-1))

We are told this equals 1/3, so 30/(n(n-1)) = 1/3

Multiply both sides by 3n(n-1):     90 = n(n-1)   or  90 = n^2 - n

Subtract 90 from both sides:    n^2 - n - 90 = 0

This factors as  (n-10)(n+9) = 0  

Since we can't have a negative number of sweets, we must have n = 10 sweets

The mean is 26.  With the standard deviation = 4  we have:

\(\frac{1}{\sqrt{2\pi}\times 4}\int_{-\infty}^{30}e^{-\frac{1}{2}(\frac{x-26}{4})^2}dx=0.8413\)

heureka's answer is the correct one.

Rewrite the following quantities in units with SI prefixes

1. 36582472g

       \( 36582472~\text{g} \\ = 36.582472\cdot 10^6~\text{g} \qquad 1~\text{Mg} = 10^6~\text{g} \\ = 36.582472~\text{Mg} \\ \mathbf{36582472~\text{g} = 36.582472~\text{Mg} } \)

2. 0.000000452m

       \( 0.000000452~\text{m} \\ = 452\cdot 10^{-9}~\text{m} \qquad 1~\text{nm} = 10^{-9}~\text{m} \\ = 452~\text{nm} \\ \mathbf{0.000000452~\text{m} = 452~\text{nm} } \)

3. 53236V

       \( 53236~\text{V} \\ = 53.236\cdot 10^3~\text{V} \qquad 1~\text{kV} = 10^3~\text{V} \\ = 53.236~\text{kV} \\ \mathbf{53236~\text{V} = 53.236~\text{kV} } \)

4. 4.62 x 10^-3s

       \(4.62\cdot 10^{-3}~s \qquad 1~\text{ms} = 10^{-3}~s \\ \mathbf{4.62\cdot 10^{-3}~s = 4.62 ~\text{ms}}\)

2/5 + -4/15=2/5-4/15=6/15-4/15=2/15

Hallo guest !

what are the last two digits of 4^2000 ?

or \(4^{2000} \equiv x \pmod {100}\)

\(\boxed{~ 4^{11} \equiv 4 \pmod{100} ~} \)

\(\small{ \begin{array}{rcl} && 4^{2000} \pmod {100} \\ &\equiv& 4^{11\cdot 181 + 9} \pmod {100}\\ &\equiv& 4^{11\cdot 181} \cdot 4^9 \pmod {100}\\ &\equiv& (\underbrace{4^{11}}_{\equiv 4 \pmod {100}})^{181} \cdot 4^9 \pmod {100}\\ &\equiv& 4^{181}\cdot 4^9 \pmod {100}\\\\ &\equiv& 4^{11\cdot 16 + 5} \pmod {100}\\ &\equiv& 4^{11\cdot 16} \cdot 4^5 \cdot 4^9 \pmod {100}\\ &\equiv& (\underbrace{4^{11}}_{\equiv 4 \pmod {100}})^{16} \cdot 4^5 \cdot 4^9 \pmod {100}\\ &\equiv& 4^{16} \cdot 4^5 \cdot 4^9 \pmod {100}\\ &\equiv& 4^{30} \pmod {100}\\\\ &\equiv& 4^{11\cdot 2 + 8} \pmod {100}\\ &\equiv& 4^{11\cdot 2} \cdot 4^8 \pmod {100}\\ &\equiv& (\underbrace{4^{11}}_{\equiv 4 \pmod {100}})^{2} \cdot 4^8 \pmod {100}\\ &\equiv& 4^{2} \cdot 4^8 \pmod {100}\\ &\equiv& 4^{10} \pmod {100}\\ &\equiv& 76\pmod {100} \end{array} }\)

The last two digits of \(4^{2000}\) is 76

Easy Q a is 500km per hour

1. how many micrograms make 1kg?             1 kg = 1 000 000 000 µg

\(\small{ \begin{array}{rcl} 1~ \text{kg} &=& 10^3 g \\ 1~ \mu g &=& 10^{-6} g \\ \\ \hline \\ \frac{ 1~ \text{kg} }{1~ \mu g } &=& \frac{10^3 g}{ 10^{-6} g } \\ \frac{ 1~ \text{kg} }{1~ \mu g } &=& \frac{10^3}{ 10^{-6} } \\ \frac{ 1~ \text{kg} }{1~ \mu g } &=& 10^3 \cdot 10^6\\ \frac{ 1~ \text{kg} }{1~ \mu g } &=& 10^9 \\ 1~ \text{kg} &=& 10^9~ \mu g \\ \end{array} }\)

2. how many nanometers are there in 1cm?  1 cm =     10 000 000 nm

\(\small{ \begin{array}{rcl} 1~ \text{cm} &=& 10^{-2} m \\ 1~ \text{nm} &=& 10^{-9} m \\ \\ \hline \\ \frac{ 1~ \text{cm} }{1~ \text{nm}} &=& \frac{ 10^{-2} m }{ 10^{-9} m } \\ \frac{ 1~ \text{cm} }{1~ \text{nm}} &=& \frac{ 10^{-2} }{ 10^{-9} }\\ \frac{ 1~ \text{cm} }{1~ \text{nm}} &=& 10^{-2} \cdot 10^9\\ \frac{ 1~ \text{cm} }{1~ \text{nm}} &=& 10^7 \\ 1~ \text{cm} &=& 10^7~ \text{nm} \\ \end{array} }\)

No I can not blast his on this special day of Eid , we shall fight again but I may not fight with you MrTin, you disappointed me , I may not fight therefore it will be hard for you since I'll directly blast you rather than playing with you .

1. how many micrograms make 1kg?

=1×10^9 µg  (micrograms)

2. how many nanometers are there in 1cm?

=1×10^7 nm  (nanometers)

(8-3^2)^6 + (2 x 3-7)^9

\(\small{ \begin{array}{rcl} (8-3^2)^6 + (2 \cdot 3-7)^9 \\ &=& (8-9)^6 + ( 6-7)^9\\ &=& (-1)^6 + (-1)^9 \\ &=& (-1)^6 [ 1 + (-1)^3] \qquad | \qquad [ 1 + (-1)^3] = [ 1 + (-1)(-1)(-1)]=[1-1]=0\\ &=& (-1)^6 \cdot 0 \\ (8-3^2)^6 + (2 \cdot 3-7)^9 &=& 0 \end{array} }\)

(8-3^2)^6 + (2 x 3-7)^9=0

/4=100

n/4=100

n=4 X 100

n=400

V²=U²+2AS,

WHERE V=FINAL VELOCITY

U=INITIAL VELOCITY

A=ACCELERATION

S=DISTANCE OF ACCELERATION

You maybe able to use this formula to solve your problem.

Hint: convert 1Km into meters.

2(-0.8+12)-6(-1.6-3)

=50