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If     x  =  t + 3     and     y  =  2t² - 4

then solve the x-equation for t:  x  =  t + 3     --->     t  =  x - 3

and substitute this into the other equation:  y  =  2t² - 4     --->     y  =  2(x - 3)² - 4

                                                                                                      y  =  2(x² - 6x + 9) - 4

                                                                                                      y  =  2x² - 12x + 18 - 4

                                                                                                      y  =  2x² - 12x + 14

https://web2.0calc.com/questions/helpsssss

 (55.5 x 2)  5 + 13 - 7

(111)(5) + 13 - 7

555 + 13 - 7

568 - 7

561

2y = -x + 3    or    y = -1/2 x + 3/2

                            y = -5x - 1

where they intersect they are equal....so equate them

-1/2 x + 3/2 = -5x-1

4 1/2 x = - 5/2

x =   -5/2  /  4 1/2    = -5/2 * 2/9  = -5/9      this is 'x'     put this value in one of the equations to compute 'y'

If 5x - 17 = -x + 7          add 17 to both sides of the  'quation

    5x = -x + 7 + 17        add 'x' to both sides

    6x = 24                     divide both sides by 6

       x = 4

Hit the   2^nd  button upper left....then the cos   button turns into  'acos'  which is inverse cosine

Why did you do A+B instead of B-A?

[m*n+97] / (n+1) =90, 

[m*n+73] / (n+1) =87, solve for n, m

n = 7 Number of tests before the last one

m = 89 - His average before the 8th and last test.

Area of a triangle =  1/2 b * h            b = base =20 ft    h = height = 16 ft

W = .2 B

W/.2 = B

  or    B = 5W      5 is 500%      this situation comes up often when dealing with % on this forum

                               while Boss's salary is 500% of Werner's it would take a 400% INCREASE from his present salary to match B

                                  i.e.    W + 400% * W = B

                                                5W = B

3(y+41)=3y+123

3y + 123 = 3y + 123       ANY value of y will be a 'solution'    ....  infinite answers

I think the largest n = 3.

[17^9 - 9^9] / 2^3 =14,775,057,001

Thanks!

\((\frac{1}{2})^3=\frac{1^3}{2^3}=\frac{1}{8}\)

I think the GCD of all elements of S = 1

x=5/2

try using a graphing calculator and graph both equations, and find the points where they intersect. 

It just simple mistake in counting the above list. It appears there are only 88 such numbers.

 [ (2y+1)(y+2)][   1/(y+2)  - 4/(2y+1) + (3y-4)/(2y^2+5y+2)]  =

2y+1  -    4y-8   +    3y-4    = 0

y -11=0

y=11

we note that \((y+2)(2y+1) = 2y^2 + 5y +2\), so we plug that in and solve for y:

\(\begin{align*} \frac{1}{y+2}-\frac{4}{2y+1} + \frac{3y-4}{2y^2 + 5y + 2}&= 0 \\ \frac{2y+1}{2y^2 + 5y + 2} - \frac{4(y+2)}{2y^2 + 5y + 2} + \frac{3y-4}{2y^2 + 5y + 2}&= 0 \\ \frac{2y+1-4y-8+3y-4}{2y^2 + 5y + 2}&=0 \\ \frac{y-11}{2y^2 + 5y + 2}&=0 \\ y-11&=0 \\ y&=\boxed{11} \end{align*} \)

note: since we multiply by \(2y^2 + 5y +2\), we have \(y \neq -2, -\frac{1}{2}\), but since \(y+2\) and \(2y+1\) are already denominators, we don't have to worry about fractions over zero.

if the points were to form a rectangle, the slopes would have to be perpindicular, or be negative reciprocals of eachother. we find the slopes of all the lines:

\(\begin{align*} \overline{AB} &\Rightarrow \frac{4-1}{0+3} = \frac{3}{3} =1 \\ \overline{BC} &\Rightarrow \frac{4-2}{0-0} = \text{vertical line} \\ \overline{CD} &\Rightarrow \frac{2+1}{0+3} = \frac{3}{3} =1 \\ \overline{DA} &\Rightarrow \frac{-1-1}{-3+3} = \text{vertical line} \\ \end{align*}\)

in order for the lines to be perpindicular, the slopes of \(\overline{AB}\) and \(\overline{CD}\) would have to be 0, which would make a straight line, instead of 1, so the points do not make a rectangle. however, we have two pairs of parallel sides, so we know we have a parallelogram.