Rom

 25=10(log(i)+16)

\(25=10\left(\log(i)+16\right) \\ \dfrac 5 2 = \log(i)+16 \\ -\dfrac {27}{2}=\log(i) \\ \mbox{Assuming we are using logs base 10 we have } \\ i = 10^{-\frac {27}{2}}\) 

A university claims that 65% of its basketball players get degrees. An investigation examines the fate of all 25 players who entered the program over a period of several years. The number of athletes who graduate is B(25, 0.65). Use the binomial probability formula to find the probability that all 25 graduate.

\(B_{25,0.65}(25)=\begin{pmatrix}25 \\25\end{pmatrix}(0.65)^{25}(0.35)^0 = (0.65)^{25}\approx 2.1\times10^{-5}\) 

bout tree fitty

Divide and simplify the following expression.

 

x^2-49/x ÷ x+7/x-7

 

\(\dfrac{\frac{x^2-49}{x}}{\frac{x+7}{x-7}}=\dfrac{x^2-49}{x}\dfrac{x-7}{x+7}=\) 

\(\dfrac{(x-7)(x+7)(x-7)}{x(x+7)}=\dfrac{(x-7)^2}{x}\)

Hi All how do you rationalise and symplify 2 root 3 over root 7

\(\dfrac {2\sqrt{3}}{\sqrt{7}} = \dfrac {2\sqrt{3}}{\sqrt{7}} \dfrac{\sqrt{7}}{\sqrt{7}} = \dfrac {2\sqrt{21}}{7}\) 

\(2.56 = \dfrac {256}{100} \mbox{ is the ratio of two integers so it is a rational number.} \\ -\dfrac{22}{9} \mbox{ is also the ratio of two integers and so also a rational number.}\)

\(\sin(\theta) < 0 \mbox{ and }\cos(\theta)>0\)

\(\sin(\theta) = -\dfrac 2 5 \Rightarrow \cos(\theta) = \pm \sqrt{1-\sin^2(\theta)}\\ \mbox{We know }\cos(\theta)>0 \mbox{ so we select the positive value.}\\ \cos(\theta) = \sqrt{1-\dfrac{4}{25}} = \dfrac{\sqrt{21}}{5}\)

\(\sec(\theta)=\dfrac 1 {\cos(\theta)} = \dfrac {5}{\sqrt{21}}= \dfrac{5\sqrt{21}}{21}\)

0.31104

\(\dfrac 1 9 z - 5 = \dfrac {7}{18} \\ \dfrac 1 9 z = \dfrac 7{18}+5 = \dfrac{7}{18}+\dfrac{90}{18} =\dfrac{97}{18}\\ z = 9\times \dfrac{97}{18} = \dfrac{97}{2}\\ z=\dfrac{97}{2}\)

Ok I'll take a stab at this.

VT ratio of 700 means that you are reading 1/700 of the actual voltage.

CT ratio of 200 means you are reading 1/200 of the actual current

V = (700)117 = 81900V

A = 200(2.5) = 500A

VA=81900 x 500 VA = 40950 kVA

\(kWatts = PF \times kVA = 0.89 \times 40950 ~kVA = 36445.5~ kWatts\\ kVAR = \sqrt{1-PF^2} \times kVA = 0.46 40950 ~kVA = 18671.6~ kVAR\)