Answer:
Line 1: m = 2
Line 2: m = 2
The lines are parallel.
Step-by-step explanation:
First, ensure both lines are in the slope-intercept form given as y = mx + b. where m is the slope of the line.
If the slope of both lines are the same, they are parallel.
If the slope of one is the negative reciprocal of the other, they are perpendicular.
If the slope of both lines are different and one is neither the reciprocal of the other, then they are neither parallel nor perpendicular.
✍️Line 1, y = 2x + 5, is already in the slope-intercept form.
✅The slope of Line 1 is 2
✍️Line 2, y - 3 = 2(x + 15), is in point-slope.
We can decide to rewrite in the slope-intercept form or directly determine the slope as it is given. The slope is 2. But to be sure, let's rewrite as y = mx + b.
y - 3 = 2(x + 15)
y - 3 = 2x + 30
Add 3 to both sides
y = 2x + 33
✅As we can see, the slope of line 2 is 2.
✍️Line 1 and line 2 has the same slope of 2, therefore the lines are parallel.
The expansion of a perfect square is
In words, the square of a sum of two terms is the sum of the squares of the two terms (
and 
), plus twice the product of the two terms (
)
So, when determining if you have a perfect square trinomial, you should have two perfect squares. Note that they don't have to be the first and third term, since you can rearrange terms as you prefer.
Triangle JKL has vertices J(−2, 2) , K(−3, −4) , and L(1, −2) .
Rule: (x, y)→(x + 8, y + 1 )
J’ (-2, 2) → (-2 + 8, 2 + 1 ) → (6, 3 )
K’ (-3, -4) → (-3 + 8, -4 + 1 ) → (5, -3 )
L’ (1, -2) → (1 + 8, -2 + 1 ) → (9, -1)
J’ (6,3)
K’ (5,-3)
L’ (9,-1)
Hope this helps!
Ok so this is conic sectuion
first group x's with x's and y's with y's
then complete the squra with x's and y's
2x^2-8x+2y^2+10y+2=0
2(x^2-4x)+2(y^2+5y)+2=0
take 1/2 of linear coeficient and square
-4/2=-2, (-2)^2=4
5/2=2.5, 2.5^2=6.25
add that and negative inside
2(x^2-4x+4-4)+2(y^2+5y+6.25-6.25)+2=0
factor perfect squares
2((x-2)^2-4)+2((y+2.5)^2-6.25)+2=0
distribute
2(x-2)^2-8+2(y+2.5)^2-12.5+2=0
2(x-2)^2+2(y+2.5)^2-18.5=0
add 18.5 both sides
2(x-2)^2+2(y+2.5)^2=18.5
divide both sides by 2
(x-2)^2+(y+2.5)^2=9.25
that is a circle center (2,-2.5) with radius √9.25
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)
