Hello,
Vertices are on a line parallele at ox (y=-3)
The hyperbola is horizontal.
Equation is (x-h)²/a²- (y-k)²/b²=1
Center =middle of the vertices=((-2+6)/2,-3)=(2,-3)
(h+a,k) = (6,-3)
(h-a,k)=(-2,-3)
==>k=-3 and 2h=4 ==>h=2
==>a=6-h=6-2=4 (semi-transverse axis)
Foci: (h+c,k) ,(h-c,k)
h=2 ==>c=8-2=6
c²=a²+b²==>b²=36-4²=20
Equation is:
Answer:
first you subtract 5 from both sides
your left with 3x=11
divide by 3 on both sides
x=3.666 repeating
Step-by-step explanation:
Answer:
9x^2 -49
Step-by-step explanation:
We recognize that this is of the form
(a+b)(a-b) = a^2 - b^2
3x = a and 7 =b
(3x)^2 - 7^2
9x^2 -49
Solution:
Using Substitution Method:
-4x+7y=-5 (Equation 1)
x-3y=-5 (Equation 2)
get the value of x from Equation 2
x=3y-5 (Equation 3)
Put the value of x from Equation 3 in Equation 1
-4(3y-5)+7y=-5
-4(3y)+20+7y=-5
-12y+7y=-5-20
-5y=-25
Negative sign on both sides cancels each other
y=25/5
y=5
Putting value of y in equation 3
x=3(5)-5
x=15-5
x=10
Therefore, [x,y]=[10,5]
Using Elimination Method
-4x+7y=-5 (Equation 1)
x-3y=-5 (Equation 2)
Multiply equation 2 with -4 in order to eliminate the x term
-4(x-3y)=-5*4
-4x+12y=20 (Equation 3)
Adding Equation 1 and 3
-4x+7y=-5
-4x+12y=20
+ - = - (Change Of Sign with x and y terms)
-----------------
0x-5y = -25
-5y=-25
y=5
Substituting y’s value is Equation 1
-4x+7(5)=-5
-4x+35=-5
-4x=-40
Cancellation of negative sign on both sides
x=40/4
x=10
[x,y]=[10,5]
I believe yes, no, no, no but I am not sure
