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I suspect this should be find the limit as x tends to zero.

If so:

(3 + 2x)^5 → 3^5 + 5*3^4*2x + O(x^2) → 243 + 10*3^4*x + O(x^2)

So:

((3+ 2x)^5 - 243)/(3x) → 10*3^3 + O(x) → 270 + O(x)

In the limit as x tends to zero, this tends to 270

.

Find the following limit:
lim_(x->3) ((2 x+3)^5-243)/(3 x)

((2 x+3)^5-243)/(3 x) = 2/3 (16 x^4+120 x^3+360 x^2+540 x+405):
lim_(x->3) 2/3 (16 x^4+120 x^3+360 x^2+540 x+405)

lim_(x->3) 2/3 (16 x^4+120 x^3+360 x^2+540 x+405) = 2/3 (405+540 3+360 3^2+120 3^3+16 3^4) = 6534:
Answer: |6534

limit problem x tends to 3 ((3+2x)^5-243)/(3x)

Perhaps I misunderstand what you are asking???

\(\displaystyle \lim_{x\rightarrow3}\;\;\frac{ (3+2x)^5-243}{3x}\\ =\frac{ (3+2*3)^5-243}{3*3}\\ =\frac{ 9^5-243}{9}\\ =\frac{ 9^5-9*27}{9}\\ =\frac{ 9^4-27}{1}\\ =6534\)

Suppose the polynomial p(x)=x^3+ax^2+bc+c has the property that the mean of its zeroes, the product of its zeroes, and the sum of its coefficients are all equal. If the y-intercept of the graph of y=p(x) is 2, what is b?

for this answer I will assume it is a typo

The y intercept is the constant so c=2

\(p(x)=x^3+ax^2+bx+c\\ p(x)=x^3+ax^2+bx+2\\\)

\(\mbox{Let the zeros be }\;\alpha,\;\; \beta, \;\;and \;\; \gamma\\ \alpha+\beta+\gamma=\frac{-a}{1}\quad so \\ \frac {\alpha+\beta+\gamma}{3}=\frac{-a}{3}\\ \alpha\beta\gamma=\frac{-c}{1}=-2\\ \mbox{The coefficients are 1, a and b}\\ so\\ \frac {\alpha+\beta+\gamma}{3}=\alpha\beta\gamma=1+a+b\\ \frac{-a}{3}=-2=1+a+b\\ a=6\\~\\ -2=1+6+b\\ -2=7+b\\ b=-9\\ \)

Dabae, 

Is it meant to be bc as you have entered OR is it meant to be bx   ??

Take first partial derivative with respect to x to get dz/dx ( sorry,can't write delta symbol) 

then take partial derivative of dz/dx with respect to y,that is you want

d/dy ( dz/dx)

you can also have

d/dx(dz/dy)  which is the same thing and will give you the same answer.

12 1\2+8 1\6+9 1\3=?

1) 20   2) 30

3) 40   4)50

\(\begin{array}{|rcll|} \hline && 12 \frac 1 2 +8 \frac 1 6 +9 \frac 1 3 \\ &=& 12 + \frac 1 2 +8 + \frac 1 6 +9 + \frac 1 3 \\ &=& 12 + 8 + 9 + \frac 1 2 + \frac 1 6 + \frac 1 3 \\ &=& 29 + \frac 1 2 + \frac 1 6 + \frac 1 3 \\ &=& 29 + \frac{ 1 } { 2 }\cdot \frac{ 3 } { 3 } + \frac 1 6 + \frac {1}{ 3} \cdot \frac{ 2 } { 2 }\\ &=& 29 + \frac{ 3 } { 6 } + \frac 1 6 + \frac {2}{ 6}\\ &=& 29 + \frac{ 3+1+2 } { 6 } \\ &=& 29 + \frac{ 6 } { 6 } \\ &=& 29 + 1 \\ &=& 30 \\ && \mathbf{12 \frac 1 2 +8 \frac 1 6 +9 \frac 1 3 = 30} \\ \hline \end{array} \)

unit digit of 2354 raised to the power of 1048???

\(\begin{array}{|rcll|} \hline && 2354^{1048} \pmod {10} \qquad & | \qquad 2354 \pmod{10} \equiv 4 \\ &\equiv& 4^{1048} \pmod {10} \\ &\equiv& 4^{2\cdot 524} \pmod {10} \\ &\equiv& (4^2)^{524} \pmod {10} \qquad & | \qquad 4^2 \pmod{10} \equiv 16 \pmod{10} \equiv 6 \\ &\equiv& 6^{524} \pmod {10} \qquad & | \qquad 6^n \pmod{10} \equiv 6 \\ &\equiv& 6 \pmod {10}\\ \hline \end{array} \)

Unit digit of 2354 raised to the power of 1048 is 6

\begin{array}{rll} 2x-8&=&6-5x\\

 7x&=&14\\

x&=&2

\end{array}

\(\dfrac{m}{x}+q=N\\ \dfrac{9}{2}+q=\dfrac{9}{5}\\ q = \dfrac{9}{5}-\dfrac{9}{2}\\ \;\;=\dfrac{18-45}{10}\\\;\;=\dfrac{-27}{10}\\\;\;=-2\dfrac{7}{10}\)

\(\text{If you want to make }\dfrac{a}{b} \text{ , type }\mathtt a/\mathtt b\text{ in the calculator on this website.}\)

Oops I was wrong in the last bit :P

Area of a square = (side length)²

x = atan (0.4557) \(\approx 0.4276 \text{ radians or }24.4988^{\circ} \)

Suppose \(r\) and \(s\) are the values of \(x\) that satisfy the equation \(x^2-2mx+(m^2+2m+3)=0\) for some real number \(m\). Find the minimum real value of \(r^2+s^2\).

Use the quadratic formula:

\(ax^2 + bx + c = 0 \implies x = \frac{1}{2}\left(-b \pm \sqrt{b^2 - 4ac}\right)\)

(here, \(a = 1\), \(b = -2m\) and \(c = m^2 + 2m + 3\))

to find that

\(r = m + \sqrt{-(2m + 3)}\)

and

\(s = m - \sqrt{-(2m + 3)}\).

Then \(r^2 + s^2 = 2\left(m^2 - 2m +3\right)\).

This is a concave-up parabola with vertex at \(m = 1\).

Now \(\left(r^2 + s^2\right)\big|_{m = 1} = 2\left(1 - 2 + 3\right) = 4\).

Hello max, so it's true.

\(y= \frac{x- 4\pm \sqrt{x^{2}- 8x+ 56 } }{2}\)

x^2+(x^2-10)^2  →  x^2 + x^4 -2*10*x^2 + 10^2  →  x^2 + x^4 -20x^2 + 100  →  x^4 + (1 - 20)x^2 + 100  →  x^4 - 19x^2 + 100

A metal having a work function of 1.2 eV has light whose photon energy is 3.6 eV shining on it. What is maximum possible kinetic energy of the electrons that are ejected?

A.5.4 × 10-19 J

B.3.8 x 10-19 J

C.4.9 × 10-19 J

D.5.8 x 10-19 J J

E.4.4 × 10-19 J

It takes 1.2 eV to eject the electron, so of the 3.6 eV energy from the photon \(3.6eV -1.2eV = 2.4 eV\) will go into the kinetic energy of the ejected electron. Next we convert from eV to Joules: \(1 eV = 1.6022x10^{-19} J\), so the ejected electron will have kinetic energy

\(KE = 2.4 eV = 3.8 \times10^{-19} J.\)

Choose answer B.

Find the ones digit of the largest power of 2 that divides into (2^4)!.

[...] integer Part

power of 2:

\(\begin{array}{|rcll|} \hline && \left[ \frac{2^4}{2^1} \right] + \left[ \frac{2^4}{2^2} \right] + \left[ \frac{2^4}{2^3} \right] + \left[ \frac{2^4}{2^4} \right] \\\\ &=& 2^3 + 2^2 + 2^1 + 1 \\\\ &=& 8+4+2+1 \\\\ &=& 15 \\\\ && 2^4! = 2^{15} \cdot \dots \\\\ \hline \end{array} \)

power of 3:

\(\begin{array}{|rcll|} \hline && \left[ \frac{16}{3^1} \right] + \left[ \frac{16}{3^2} \right] \\\\ &=& 5 + 1 \\\\ &=& 6 \\\\ && 2^4! = 2^{15} \cdot 3^6 \dots \\\\ \hline \end{array}\)

power of 5:

\(\begin{array}{|rcll|} \hline && \left[ \frac{16}{5^1} \right] \\\\ &=& 3 \\\\ && 2^4! = 2^{15} \cdot 3^6 \cdot 5^3 \dots \\\\ \hline \end{array} \)

power of 7:

\(\begin{array}{|rcll|} \hline && \left[ \frac{16}{7^1} \right] \\\\ &=& 2 \\\\ && 2^4! = 2^{15} \cdot 3^6 \cdot 5^3 \cdot 7^2 \dots \\\\ \hline \end{array}\)

power of 11:

\(\begin{array}{|rcll|} \hline && \left[ \frac{16}{11^1} \right] \\\\ &=& 1 \\\\ && 2^4! = 2^{15} \cdot 3^6 \cdot 5^3 \cdot 7^2 \cdot 11^1 \dots \\\\ \hline \end{array}\)

power of 13:

\(\begin{array}{|rcll|} \hline && \left[ \frac{16}{13^1} \right] \\\\ &=& 1 \\\\ && 2^4! = 2^{15} \cdot 3^6 \cdot 5^3 \cdot 7^2 \cdot 11^1 \cdot 13^1 \\\\ \hline \end{array} \)

\(16! = 2^{15} \cdot 3^6 \cdot 5^3 \cdot 7^2 \cdot 11 \cdot 13\)

Further steps on asinus's last step.

\(\huge\color{blue}y=\dfrac{x-4\pm\sqrt{(4-x)^2+40}}{2}\)

Oops is it too big xD

\(\huge\color{blue}y=\dfrac{x-4\pm\sqrt{-x^2+8x+24}}{2}\)

9.47 x 5 = 47.35

Use the calculator!! It's just located at the home page of this website.

A negative number just means 0 - x . 

e.g. : -1 = 0-1 , - 1/3 = 0 - 1/3.

You can't really 'make' a negative number.

0.545+3[0.545]^1/2

= \(0.545^{\frac{1}{2}}(0.545^{\frac{1}{2}}+3)\)

\(\approx 0.738 (3.738)\\ \text{Which is 2.758644}\)

\(\displaystyle\log_x16 = \dfrac{4}{3}\\ x^{\frac{4}{3}} = 16\\ x^4 = 16^3\\ \;\;\;\;=4096\\ x= \pm\sqrt[4]{4096}\\ \;\;=\pm 8\)

\(\text{log with base -8 is impossible} \therefore \text{x = 8}\)

Same as geno's answer