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nvm i got it answered

ok ok

It' fine! Don't worry about it. It's only 4 questions. I do have to get it done today though, it's already almost 12 and I have til 2. >.<

im sorry 'bout that i am punching myself for that

*sigh* it was a,b,c,and d thanks for your help 

I have to re-take everything >.< 

Okay but I'm pretty sure it's more than 2

yes ma'am

So...A and B?

do will's answer

the last answer k

Okay...now I don't know what to put >.<

I, II, and III

The c term basically decides how much the parabola is shifted vertically. Moving the parabola would change all three of those things.

y = +/-(sqrt(3^2 - (x+5)^2)) - 5

y = +/-(sqrt(x^2 + 10x - 16)) - 5

maybe b

A is true. The probability of winning with the first card does not affect the probability of losing with the second card.

B is true. The probability of A winning and B losing is equal to the product of their probabilities because they are independant.

I feel like it's more than just those 2

i am thinking C and D

Thank you xo much!

Let us assume the price of oil at BEGINNING of 1st full year in business =$100
The price of oil at the END of the first year in business =$100 x 1.5 =$150
The price of oil at the BEGINNING of the 2nd year in business =$150
The price of oil at the END of the second year in business =$150 x 0.5=$75
$75/$100 =0.75 x 100 =75% - percentage of oil price at the END of 2nd year compared to the price of oil at the BEGINNING of 1st.year.

20! / [16! x 4! x 2^20] = 0.00462 x 100 =o.462%

what is the probability of getting 16 out of 20 questions correct on a true/false test?

20C16*0.5^20 

nCr(20,16)*(1/2)^20 = 0.0046205520629883 = 0.46%

That is assuming that you totally guess every single answer.

so what is the correct answer?

Inverse Functions: Express your answer in degrees, minutes, seconds to the nearest hundredth of a second.

1.

\(\begin{array}{|rcll|} \hline \cos(\varphi_1) &=& 0.4444 \\ \varphi_1 &=& \arccos(0.4444) \\ \varphi_1 &=& 63.6150426705^{\circ}\\ \varphi_1 &=& 63^{\circ}\ 36^{'}\ 54.15^{''}\\ \hline \end{array}\)

2.

\(\begin{array}{|rcll|} \hline \cos(\varphi_2) &=& 0.4444 \\ \varphi_2 &=& -\arccos(0.4444) + 360^{\circ} \\ \varphi_2 &=& -63.6150426705^{\circ} + 360^{\circ} \\ \varphi_2 &=& 296.384957330^{\circ} \\ \varphi_2 &=& 296^{\circ}\ 23^{'}\ 5.85^{''}\\ \hline \end{array}\)

Consider the function g(x)=3x-4. 

For what value of  a  is g(a)=0 ?

\(g(a) = 0:\)

\(\begin{array}{|rcll|} \hline g(a)= 0 &=&3a-4 \\ 0&=& 3a-4 \quad & | \quad +4 \\ 4&=& 3a \quad & | \quad :3 \\ \dfrac{4}{3} &=& a \\ \mathbf{ a } &\mathbf{=}& \mathbf{ \dfrac{4}{3} } \\ \hline \end{array} \)