<span>3n + 4 = 7
</span><span>3n = 7 - 4
3n = 3
n = 1</span>
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
Answer:
.
Step-by-step explanation:
If (α, β) are the coordinates of the center of the hyperbola, then its equation of the hyperbola is 
.
Now, the vertices of the hyperbola are given by (α ± a, β) ≡ (1,-3) and (-3,-3)
Hence, β = - 3 and α + a = 1 and α - a = -3
Now, solving those two equations of α and a we get,
2α = - 2, ⇒ α = -1 and
a = 1 - α = 2.
Now, eccentricity of the hyperbola is given by ![b^{2} = a^{2}(e^{2} - 1) = 4[(\frac{5}{2} )^{2} -1] = 21](https://tex.z-dn.net/?f=b%5E%7B2%7D%20%3D%20a%5E%7B2%7D%28e%5E%7B2%7D%20%20-%201%29%20%3D%204%5B%28%5Cfrac%7B5%7D%7B2%7D%20%29%5E%7B2%7D%20-1%5D%20%3D%2021)
{Since 
given}
Therefore, the equation of the given hyperbola will be
. (Answer)
