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Let the height of the shorter building be h metres and the width of the street be w metres.

tan(38°) =  (h + 30)/w              (1)

tan(23°) =  h/w                        (2)

Subtract (2) from (1) to get    tan(38°) - tan(23°) = 30/w   so  w = 30/(tan(38°) - tan(23°))

I'll leave you to crunch the numbers.

4 persons 4! = 24

The cucumfrence of a circle divided by the diamiter=pi-3.14

I need to get a formula for "x" from:

\(\small{ f=nw+\frac{l}{d}(d-2x-z)+m-\sqrt{w} }\)

Please help :(

\(\small{ \begin{array}{rcll} f &=& nw+\frac{l}{d}(d-2x-z)+m-\sqrt{w} & | \quad +\sqrt{w} \\ f +\sqrt{w} &=& nw+\frac{l}{d}(d-2x-z)+m & | \quad -m \\ f +\sqrt{w} - m &=& nw+\frac{l}{d}(d-2x-z) & | \quad -nw \\ f +\sqrt{w} - m -nw &=& \frac{l}{d}(d-2x-z) & | \quad \cdot \frac{d}{l} \\ ( f +\sqrt{w} - m -nw )\cdot \frac{d}{l} &=& d-2x-z & | \quad +z \\ ( f +\sqrt{w} - m -nw )\cdot \frac{d}{l}+z &=& d-2x & | \quad -d \\ ( f +\sqrt{w} - m -nw )\cdot \frac{d}{l}+z-d &=& -2x & | \quad :(-2) \\ \frac{ ( f +\sqrt{w} - m -nw )\cdot \frac{d}{l}+z-d }{-2} &=& x \\ x &=& \frac{ ( f +\sqrt{w} - m -nw )\cdot \frac{d}{l}+z-d }{-2}\\ x &=& -\frac{ ( f +\sqrt{w} - m -nw )\cdot \frac{d}{l}+z-d }{2}\\ x &=& \frac{ -( f +\sqrt{w} - m -nw )\cdot \frac{d}{l}-z+d }{2}\\ \mathbf{x} &\mathbf{=}& \mathbf{\frac{ d-z-( f +\sqrt{w} - m -nw )\cdot \frac{d}{l} }{2} } \end{array} }\)

the following system of equations is shown below. 2x+3y=4 x - y = -3 which of the below is the solution to this system? A) (-13,10) B) (-1,2) C) (1,-2) D) (2,3)

\(\small{ \begin{array}{lrcll} (1) & 2x+3y &=& 4 \qquad & | \qquad -2x\\ & 3y &=& 4-2x \qquad & | \qquad :3\\ & y &=& \frac43-\frac23x \\\\ (2) & x - y &=& -3 \qquad & | \qquad \cdot (-1)\\ & -x + y &=& 3 \qquad & | \qquad +x\\ & y &=& 3+x \\ \hline \\ (2) = (1) \\ & y = 3+x &=& \frac43-\frac23x \\ & 3+x &=& \frac43-\frac23x \qquad & | \qquad +\frac23x\\ & 3+x+\frac23x &=& \frac43 \qquad & | \qquad -3\\ & x+\frac23x &=& \frac43 -3\\ & \frac33x+\frac23x &=& \frac43 -\frac33\cdot 3\\ & \frac33x+\frac23x &=& \frac43 -\frac93\\ & \frac53x &=& -\frac53 \qquad & | \qquad \cdot 3\\ & 5x &=& -5 \qquad & | \qquad :5\\ & x &=& -\frac55 \\ & \mathbf{ x } & \mathbf{=} & \mathbf{-1} \\\\ & y &=& 3+x \\ & y &=& 3-1 \\ & \mathbf{ y } & \mathbf{=} & \mathbf{2} \\\\ \end{array} }\)

B) (-1,2) is the answer

6 times 1/2 is the same as saying 6 divided by 2.

Meaning the quotient/product (however you go about this) would be 3.

what does e mean in math?

The number e is an important mathematical constant that is the base of the natural logarithm.

It is approximately equal to  2,71828 18284 59045 23536 02874 71352 66249 77572 47093 69995 …

Sometimes called Euler's number after the Swiss mathematician Leonhard Euler.

Also like \(\pi\), e is transcendental

Determine if the two equations are parallel, perpendicular, or the same line.  If, perpendicular, at what point does the two lines intersect?

\(\small{ \begin{array}{lrclll} (1) & 2x+3y = -2 \qquad \Rightarrow \qquad y &=& -\frac23 x - \frac23 & m_1 = -\frac23 & b_1 = -\frac23 \\ (2) & x+y= 0 \qquad \Rightarrow \qquad y &=& -x & m_2 = -1 & b_2 = 0 \\ \end{array} }\)

1. Parallel ?

Parallel, if  \(\small{m_1 = m_2 \text{ and } b_1 \ne b_2}\)

We have: \(\small{m_1 \ne m_2 \qquad -\frac23 \ne -1 } \qquad \Rightarrow {\color{red}parallel~ no} \)

2. The same line ?

The same line, if  \(\small{m_1 = m_2 \text{ and } b_1 = b_2}\) 

We have:

\(\small{m_1 \ne m_2 \qquad  -\frac23 \ne -1 } \qquad \Rightarrow {\color{red}the~ same~ line~ no}\)\(\small{b_1 \ne b_2 \qquad  -\frac23 \ne 0 } \qquad \Rightarrow {\color{red}the~ same~ line~ no}\) 

3. Perpendicular ?  

Perpendicular, if  \(\small{m_1 = -\frac{1}{m_2}}\)  

We have: \(\small{m_1 \ne -\frac{1}{m_2} \qquad  -\frac23 \ne -\frac{1}{-1} } \qquad \Rightarrow {\color{red}perpendicular~ no}\)

Every job available uses math either directly or indirectly.

-2 / (2/5) = x.

-2/1 = 2/5x

Cross Multiply.

-10 = 2x

Division property of equality.

-5 = x

Your final answer is -5.

\(a=\pi\times{r}^{2}\)

\(a=area\)

\(\pi=pi\)

\(r=radius\)

There are three ways to solve this expression

Way One:

\(\frac{14}{6}\times\frac{24}{7}\)

\(\frac{336}{42}\)

\(8\)

Way Two:

\(\frac{14}{6}\times\frac{24}{7}\)

\(\frac{7}{3}\times\frac{24}{7}\)

\(\frac{168}{21}\)

\(8\)

Way Three:

\(\frac{14}{6}\times\frac{24}{7}\)

\(\frac{7}{3}\times\frac{24}{7}\)

\(\frac{1}{3}\times24\)

\(\frac{24}{3}\)

\(8\)

Yes, the way you have them, they cancel each other to zero.

I'm going on the assumption that "x2" means \({x}^{2}\)

We start with:

28=x^2-3x

If you want to find solutions then you need to have the equation equal zero:

0=x^2-3x-28

If you factor then you get:

0=(x+4)(x-7)

Solve each binomial as if it were a separate problem

x+4=0   and   x-7=0

Meaning... 

x=-4 and 7

∞.  On my MacBook Pro I hold down the option key and then push the 5 key.  On a PC, hold down the ALT key and then push the 2 key, then the 3 key, and then the 6 key all on the num-lock keypad.