First, you have to get the equation to the normal form of a line. To do this, you divide both sides by -6 to get
. The y-intercept is at -4 and your slope is 2/3. Use this to graph the equation.
1/3 x 4/4 = 4/12
4/12 + 5/12 = 9/12
12/12 - 9/12 = 3/12
3/12 = 1/4
(xy)' + (2x)' + (3x^2)' = (4)'
y + xy' + 2 + 6x = 0
xy' = -y -2 -6x
y' = [-y -2 -6x] / x
Now solve y from the original equation and substitue
xy + 2x + 3x^2 = 4 => y = [-2x - 3x^2 + 4] / x
y' = [(-2x - 3x^2 +4) / x - 2 - 6x ] / x
y' = [-2x - 3x^2 + 4 -2x -6x^2 ] x^2 = [ -4x - 9x^2 + 4] / x^2 =
= [-9x^2 - 4x + 4] / x^2
First one is 4/ 52 and the second second one is 4/51. (I did not reduce).
9514 1404 393
Answer:
yes
Step-by-step explanation:
The triangles are given as right triangles. Hypotenuses QT and RS are given as congruent.
We also have XS ≅ TP. By the addition property of equality, this means ...
XS +ST ≅ ST +TP
By the segment sum theorem, this means ...
XT ≅ SP
XT and SP are the corresponding legs of the right triangles. So, we have corresponding hypotenuses and corresponding legs congruent. This lets us conclude ΔXQT ≅ ΔPRS by the HL theorem.
