Step-by-step explanation:
1 Remove parentheses.
8{y}^{2}\times -3{x}^{2}{y}^{2}\times \frac{2}{3}x{y}^{4}
8y
2
×−3x
2
y
2
×
3
2
xy
4
2 Use this rule: \frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}
b
a
×
d
c
=
bd
ac
.
\frac{8{y}^{2}\times -3{x}^{2}{y}^{2}\times 2x{y}^{4}}{3}
3
8y
2
×−3x
2
y
2
×2xy
4
3 Take out the constants.
\frac{(8\times -3\times 2){y}^{2}{y}^{2}{y}^{4}{x}^{2}x}{3}
3
(8×−3×2)y
2
y
2
y
4
x
2
x
4 Simplify 8\times -38×−3 to -24−24.
\frac{(-24\times 2){y}^{2}{y}^{2}{y}^{4}{x}^{2}x}{3}
3
(−24×2)y
2
y
2
y
4
x
2
x
5 Simplify -24\times 2−24×2 to -48−48.
\frac{-48{y}^{2}{y}^{2}{y}^{4}{x}^{2}x}{3}
3
−48y
2
y
2
y
4
x
2
x
6 Use Product Rule: {x}^{a}{x}^{b}={x}^{a+b}x
a
x
b
=x
a+b
.
\frac{-48{y}^{2+2+4}{x}^{2+1}}{3}
3
−48y
2+2+4
x
2+1
7 Simplify 2+22+2 to 44.
\frac{-48{y}^{4+4}{x}^{2+1}}{3}
3
−48y
4+4
x
2+1
8 Simplify 4+44+4 to 88.
\frac{-48{y}^{8}{x}^{2+1}}{3}
3
−48y
8
x
2+1
9 Simplify 2+12+1 to 33.
\frac{-48{y}^{8}{x}^{3}}{3}
3
−48y
8
x
3
10 Move the negative sign to the left.
-\frac{48{y}^{8}{x}^{3}}{3}
−
3
48y
8
x
3
11 Simplify \frac{48{y}^{8}{x}^{3}}{3}
3
48y
8
x
3
to 16{y}^{8}{x}^{3}16y
8
x
3
.
-16{y}^{8}{x}^{3}
−16y
8
x
3
Done
Hello!
To find the equation of a line parallel to y = 3x - 3 and passing through the point (4, 15), we need to know that if two lines are parallel, then their slopes are equivalent.
This means that we create a new equation in slope-intercept form, which includes the original slope, which is equal to 3.
In slope-intercept form, we need a y-intercept. So, we would substitute the given ordered pair into the new equation with the same slope and solve.
Remember that slope-intercept form is: y = mx + b, where m is the slope and b is the y-intercept.
y = 3x + b (substitute the ordered pair (4, 15))
15 = 3(4) + b (simplify)
15 = 12 + b (subtract 12 from both sides)
3 = b
Therefore, the equation for the line parallel to the line y = 3x - 3, and passing through the point (4, 15) is y = 3x + 3.
9 sq. in because each side is 3 inch and area is length x width. each side is e inch so 3x3=9 sq in.
Multiply the original DE by xy:
xy2(1+x2y4+1−−−−−−−√)dx+2x2ydy=0(1)
Let v=xy2, so that dv=y2dx+2xydy. Then (1) becomes
x(y2dx+2xydy)+xy2x2y4+1−−−−−−−√dxxdv+vv2+1−−−−−√dx=0=0
This final equation is easily recognized as separable:
dxxln|x|+CKxvKx2y2−1K2x4y4−2Kx2y2y2=−dvvv2+1−−−−−√=ln∣∣∣v2+1−−−−−√+1v∣∣∣=v2+1−−−−−√+1=x2y4+1−−−−−−−√=x2y4=2KK2x2−1integrate both sides
They are similar because they both are triangles. But they are not the same size.
