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:
Alright, let's do all of these (though this is a bit long).
1.
The constant is 1.8. All other values are coefficients to variables, which as the name implies will change.
2.
1 hour is 60 minutes, 1 minute is 60 seconds.
So, 4.2 *60 *60 = 15120 seconds.
3.
<span>−5x−4(x−6)=−3-5x-4(x-6)=-3
Let's move all x to one side, and all other numbers to another.
-5x-4(x-6)=-3-5x-4(x-6)=-3
x can be any value you want, if you actually solve this you'll only end up with -3 = -3, which is correct, of course.
Let me show you:
</span><span>−5x−4(x−6)=−3-5x-4(x-6)=-3
+5x +4(x-6) +5x +4(x-6)
-3 = -3
The value of x is irrelevant, then. X can be any real number.
4.
I'm going to assume it was an error in printing with this? If not please correct me.
m=a+2b(or b2)
subtract 2b from each
a=m-2b
(This question seems kind of odd. We should probably address this in the comments.)
5.
</span><span>5(x−2)<−3x+6
Move all x to one side, numbers to other.
5x-10<-3x+6
+3x +3x
+10 +10
8x<16
/8
<span>x < 2
</span>6.
y-3=3(x-5)
alright, to find zeros set one variable to zero and solve
x first
-3=3x-15
+15 +15
3x=12
/3
x=4
x-int is (4,0)
now y
</span>y-3=3(0-5)
y-3=-15
+3 +3
y=-12
so y-int is (0,-12)
i've got to sleep now so i'll do the rest tomorrow. Sorry for the incomplete answer.
<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em>⤴</em><em>⤴</em><em>⤴</em>
<em>Hope</em><em> </em><em>this</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em>
Answer:
-2
Step-by-step explanation:
i solved it
you gotta do pemdas
In the terms it’s none of them because this is way to confusing