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\(\text{Circumference} = \pi (\text{Diameter})\\ \text{Circumference} = 13\pi \text{ cm}\)

\(9x^2-18x-16=0\\ (3x-8)(3x+2) = 0\\ x = \dfrac{8}{3}\text{ OR }x = \dfrac{-2}{3}\)

\(15x^2+28x+12 = 0\\ (3x+2)(5x+6) = 0\\ x = \dfrac{-2}{3} \text{ OR } x = \dfrac{-6}{5}\)

Only x = -2/3 satisfies both equations. Therefore the answer is -2/3.

\((x-p)(x-q) = x^2+4x+6\\ (x-p)(x-q) = x^2+bx+c\\ \text{Comparing coefficients,} \\ (b,c) = (4,6)\\ \therefore b + c = 10\)

5)

\(\begin{cases} x+y=3\\ x^2+y^2=6\\ x^4=y^4+18\sqrt3 \end{cases}\\ (x + y)^2 = 9\\ x^2 + y^2 + 2xy = 9\\ 2xy = 3\\ x = \dfrac{3}{2y}\\ \dfrac{3}{2y} + y = 3\\ 2y^2 - 6y + 3 = 0\\ y = \dfrac{6\pm2\sqrt{3}}{4} = \dfrac{3\pm\sqrt{3}}{2}\\ \text{When } y = \dfrac{3+\sqrt3}{2}\text{,}\\ x = \dfrac{3-\sqrt3}{2}\\ \text{When } y = \dfrac{3-\sqrt3}{2}\text{,}\\ x = \dfrac{3+\sqrt3}{2}\\ \therefore (x,y) = \left(\dfrac{3+\sqrt3}{2},\dfrac{3-\sqrt3}{2}\right) \text{ OR }\left(\dfrac{3-\sqrt3}{2},\dfrac{3+\sqrt3}{2}\right)\)

4)

\(m+n = \dfrac{-5}{4}\\ mn = \dfrac{3}{4}\\ (m+7)(n+7) = mn + 7(m+n) + 49 = \dfrac{3}{4}+7\left(\dfrac{-5}{4}\right) + 49 = 41\)

3)

\(s+t = -\dfrac{9}{4}\\ st = \dfrac{-6}{4} = -\dfrac{3}{2}\\ \dfrac{s}{t} + \dfrac{t}{s} = \dfrac{(s+t)^2 - 2st}{st} = \dfrac{\left(\frac{-9}{4}\right)^2 -2\left(\frac{-3}{2}\right)}{\frac{-3}{2}} =\dfrac{-43}{8}\)

Thank you so much

2) 

\(a+b = -\dfrac{-8}{2} = 4\\ ab = \dfrac{7}{2}\\ \dfrac{1}{2a} + \dfrac{1}{2b} = \dfrac{1}{2}\left(\dfrac{a+b}{ab}\right) = \dfrac{1}{2}\left(\dfrac{4}{\frac{7}{2}}\right)=\dfrac{4}{7}\)

1)

\(d+e = \dfrac{-10}{3}\\ de = \dfrac{-25}{3}\\ (d-e)^2 = (d+e)^2 - 4de = \left(\dfrac{-10}{3}\right)^2 - 4\left(\dfrac{-25}{3}\right) = \dfrac{400}{9}\)

Calculate \(\displaystyle \prod^{2^{2019}-1}_{k=2}\log_k(k+1)\).

\(\quad (d - e)^2\\ = d^2 + e^2 - 2de\\ =(d+e)^2 - 4de\\ =\left(\dfrac{-10}{3}\right)^2 - 4\left(\dfrac{-25}{3}\right)\\ =\dfrac{100}{9} + \dfrac{100}{3}\\ =\dfrac{400}{9}\)

Compute the sum of all the solutions of \((3x+1)(x-7)+(x-3)(3x+1)=0.\)Express your answer as a fraction.

Guest:

Do you know the concept of inverse functions? That works because the exponential function and the corresponding logarithmic function are inverse functions of each other. For instance, \(f(x) = 10^x\) and \(g(x) = f^{-1}(x) = \log_{10}x\) are inverse functions and \(f(x) = 2^x\) and \(g(x) = f^{-1}(x) = \log_{2}x\). A property of inverse functions is that \(f(f^{-1}(x)) = f^{-1}(f(x)) = x\). That means \(\log_{10}(10^x) = x\) and \(10^{\log_{10} x} = x\). That's the mechanism behind how it works.

4/16=1/4

1/4*717693=717693/4

717693/4=179423.25

HOPE THIS HELPS!

WOW, I did not think bar codes whould ahve been thought of back then!

Thanks for enlightening us Ginger. :)

This is very simple! The reasons you do not know the formula is because you are not paying attention in class and you are not reading your assigned texts.  Translation: you are fucking around instead of studying.

This question describes a class-three (3) fulcrum. The fulcrum is to one side (the broom top), the load is on the other side, and the effort is in the middle. A class-three (3) lever always has an Ideal Mechanical Advantage of less than 1. For a broom, the mechanical disadvantage translates to a distance advantage. That is, a short motion near the fulcrum results in a longer motion further away.  

The formula is the same for all three classes of levers.

  \(\bf IMA = \dfrac{d_i}{d_o}\) where i and o are the input and output distances from the fulcrum. 

GA

CPhill,

You may appreciate the first “Mad Magazine” with a UPC.

Despite Mad’s hope of jamming every computer, in 1978 it’s unlikely that even one retail sale was preceded by a scan of this UPC. 

GA

No.

$350(1-0.025) - 700(0.025) = $341.25 - $17.5 = $323.75

\(deg(p(x)) > deg(q(x)) \Rightarrow deg((p+q)(x)) = deg(p(x))\)

\(c_0 + c_1 + c_2 + \dots + c_n = f(1) =32\)

Per the table, a 2.6% loan for 4 years (i.e., 48 months) will cost her 21.958 dollars per month for every thousand she finances.  So how many thousands will she need to borrow?  The car costs 14,390, but with a 1,000 trade in and 1,500 in cash, she ends up needing to borrow 11,890. 

11,890 can be written as 11.890 x 1,000 so Elise's monthly payment will be 11.890 x 21.958 

A.  She will borrow 14,390 minus 1,000 minus 1,500 = 11,890 dollars

B.  Her monthly payment will be 11.890 x 21.958 = 261.08 dollars per month

The problem didn't ask for the following, but here it is anyway.

C.  For the 11,890 she borrows, she will end up paying back a total of 48 x 261.08 = 12,531.84 

.

\(\text{There are 2 T's}\\ \text{So let's look at permutations by how many T's they contain}\\ 0 - \dbinom{4}{0}\dbinom{8}{4}4! = 1680\\ 1 - \dbinom{4}{1}\dbinom{8}{3}3! = 1344 \\ 2 -\dbinom{4}{2}\dbinom{8}{2}2! = 336\\ \text{These sum to }n = 3360\)

7*3 - 15/5  =

21 - 3  =

18

60000 =  A ( 1 + .07/ 4)^(5 * 4)     where A is the amount we need

60000 = A ( 1 + .0175)^20

60000 = A ( 1.0175)^20         diide both sides by  (1.0175)^20

60000 / ( 1.0175)^20   = A  ≈  $42,409.47

Post your questions one at a time and seperately.   

Think before you post the next and discuss what you have done (more thoroughly)

Plot them on the desmos calculator that CPhil has given you the address of.  Maybe then you can answer yourself.