Make an equation where
Y =
(any number not zero) times X² + (any number at all) times X + (any number at all)
The graph of that equation is guaranteed to be a parabola.
From the solution of the expression, it can be seen that s = 6 when t = 3 while t = 1 when s = 2.
<h3>How do we solve a mathematical expression?</h3>
Given:
(12t) = (6s) ........................ (1)
When t = 3, we can solve for s from the expression in equation (1) by substituting t = 3 into the equation as follows:
12 * 3 = 6s
36 = 6s
s = 36 / 6
s = 6
When s = 2, we can solve for t from the expression in equation (1) by substituting s = 2 into the equation as follows:
12t = 6 * 2
12t = 12
t = 12 / 12
t = 1
Learn more about mathematical expression here: brainly.com/question/12401681.
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3y - 12 = -6
3y = 6 | Add 12 to both sides
y = 2 | Divide by 3 on both sides
Answer: y = 2 (Answer B)
Answer:
Step-by-step explanation:
Okay, so I think I know what the equations are, but I might have misinterpreted them because of the syntax- I think when you ask a question you can use the symbols tool to input it in a more clear way, otherwise you can use parentheses and such.
Problem 1:
(x²)/4 +y²= 1
y= x+1
*substitute for y*
Now we have a one-variable equation we can solve-
x²/4 + (x+1)² = 1
x²/4 + (x+1)(x+1)= 1
x²/4 + x²+2x+1= 1
*subtract 1 from both sides to set equal to 0*
x²/4 +x^2+2x=0
x²/4 can also be 1/4 * x²
1/4 * x² +1*x² +2x = 0
*combine like terms*
5/4 * x^2+2x+ 0 =0
now, you can use the quadratic equation to solve for x
a= 5/4
b= 2
c=0
the syntax on this will be rough, but I'll do my best...
x= (-b ± √(b²-4ac))/(2a)
x= (-2 ±√(2²-4*(5/4)*(0))/(2*(5/4))
x= (-2 ±√(4-0))/(2.5)
x= (-2±2)/2.5
x will have 2 answers because of ±
x= 0 or x= 1.6
now plug that back into one of the equations and solve.
y= 0+1 = 1
y= 1.6+1= 2.6
Hopefully this explanation was enough to help you solve problem 2.
Problem 2:
x² + y² -16y +39= 0
y²- x² -9= 0
Please consider the attached file.
We can see that triangle JKM is a right triangle, with right angle at M. Segment KM is 6 units and segment MJ is 3 units. We can also see that KJ is hypotenuse of right triangle.
We will use Pythagoras theorem to solve for KJ as:
Now we will take positive square root on both sides:
Therefore, the length of line segment KJ is 
and option D is the correct choice.
