Judging by the question at hand I generated this equation.
x+y=12
x=2y
I begin this question by plugging in the x=2y into the equation for x.
So the new equation should be 3y=12. I then divide the entire equation by 3 to get y=4.
Next I plug y=4 into the equation, the new equation should be x+4=12. I then subtract 4 from both sides to get x=8.
The two numbers are :
x=8 y=4
Answer:
see explanation
Step-by-step explanation:
Calculate C by adding corresponding components of A + B
C =
+ ![\left[\begin{array}{ccc}-2.5\\5\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2.5%5C%5C5%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}4-2.5\\-7.5+5\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4-2.5%5C%5C-7.5%2B5%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}1.5\\-2.5\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1.5%5C%5C-2.5%5C%5C%5Cend%7Barray%7D%5Cright%5D)
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: the variable is p=6 .
Step-by-step explanation:
Isolate the variable by dividing each side by factors that don't contain the variable
The first term (a) is - 18
You add 5 to get to the next term. Or you can solve it by taking any 2 consecutive terms and find their difference.
Formula
d = t4- t3
Givens
t4 = - 3
t3 = - 8
Solution 1
d = t4 - t3 Substitute
d = -3 - ( - 8) Remove the brackets
d = -3 + 8 Combine
d = 5 Difference
Remark
Find the general formula
tn = - a + (n - 1)d Substitute
So term 20 = Example
t20 = -18 + (20 - 1)*5 Combine the inside of the brackets. Remove the brackets
t20 = - 18 + 19*5 Combine 19 and 5
t20 = -18 + 95 "Subtract"
t20 = 77 Answer