All Answers 358181 Answers

Please Dont Post All Your Work, also if you have more than one problem post it all in one form

x=0;n=x^2+2*x+17;d=2*x+5;if(int(n/d)==x and n%d==7, goto loop, goto next);loop:printx,n,d;next:x++;if(x<100, goto1, discard=0;

x          n         d
2         25        9

Now who might you be tho? I ask thee this question.

Nvm

I got it

starting with the original expression                                                       x² + ax = –14

add 14 to both sides to get a quadratic equation                                   x² + ax + 14 = 0

we're going to factor that quadratic equation

observe that there are only two ways to get 14 ... namely 7*2 or 14*1

we could use –7 times –2 or –14 times –1 but that would cause a to be negative so we discard that idea

try them both        (x + 14)(x + 1) gives us x² + 15x +14 ... in this instance a = 15

                             (x + 7)(x +2)    gives us x² + 9x + 14 ...  in this instance a = 9

15 is larger than 9 so it looks like the answer is 15

.

You have got to be joking!

\(f(x)= {1\over 1-x^2}\\ f(x)= (1-x^2)^{-1}\\ f'(x)= -1(1-x^2)^{-2}*-2x\\ f'(x)= \frac{2x}{(1-x^2)^{2}}\\\)

A ball travels on a parabolic path in which the height (in feet) is given by the expression \(-25t^2+ 75t +24\),

where t is the time after launch.

At what time is the height of the ball at its maximum?

Let the price of 1 kg of plums =P
Let the price of 1 kg of apples =A

3P + 2A = 3.40 pounds................(1)
1P + 4A = 2.80 pounds................(2), solve for A, P
Use elimination or substitution and should get:

A =1 / 2 pound / kg
P =4 / 5 pound / kg
So, 1 kg of apples costs: 1/2 pound x 100 pence?
= 50 pence

A ball travels on a parabolic path in which the height (in feet) is given by the expression \(-25t^2+ 75t +24\),

where t is the time after launch.

At what time is the height of the ball at its maximum?

\(\begin{array}{|rcll|} \hline -25t^2+ 75t +24 &=& 0 \\ t &=& \dfrac{ -75 \pm \sqrt{75^2-4\cdot(-25)\cdot 24} }{2\cdot(-25)} \\ \hline t_1 &=& \dfrac{ -75 + \sqrt{75^2-4\cdot(-25)\cdot 24} }{-50} \\\\ t_2 &=& \dfrac{ -75 - \sqrt{75^2-4\cdot(-25)\cdot 24} }{-50} \\ \hline t_{\text{max}} &=& \dfrac{t_1+t_2}{2} \\\\ t_{\text{max}} &=& \dfrac{\dfrac{ -75 + \sqrt{75^2-4\cdot(-25)\cdot 24} }{-50} +\dfrac{ -75 - \sqrt{75^2-4\cdot(-25)\cdot 24} }{-50} }{2} \\\\ t_{\text{max}} &=& \dfrac{ -75 + \sqrt{75^2-4\cdot(-25)\cdot 24}-75 - \sqrt{75^2-4\cdot(-25)\cdot 24} } {(-50)\cdot 2} \\\\ t_{\text{max}} &=& \dfrac{ -150 } { -100 } \\\\ \mathbf{t_{\text{max}}} &=& \mathbf{1.5} \\ \hline \end{array}\)