Equation of an ellipse
→having center (0,0) , vertex ( and covertex and focus is given by:
As definition of an ellipse is that locus of all the points in a plane such that it's distance from two fixed points called focii remains constant.
Consider two points (a,0) and (-a,0) on Horizontal axis of an ellipse:
Distance from (a,0) to (c,0) is = a-c =
Distance from (-a,0) to (c,0) is = a + c =
a -c + a +c
= a + a
= 2 a →(Option A )
Answer:
x = 1/3
General Formulas and Concepts:
<u>Pre-Algebra</u>
- Order of Operations: BPEMDAS
- Equality Properties
Step-by-step explanation:
<u>Step 1: Define equation</u>
4x + 1 = 3 - 2x
<u>Step 2: Solve for </u><em><u>x</u></em>
- Add 2x on both sides: 6x + 1 = 3
- Subtract 1 on both sides: 6x = 2
- Divide 6 on both sides: x = 1/3
<u>Step 3: Check</u>
<em>Plug in x to verify it's a solution.</em>
- Substitute: 4(1/3) + 1 = 3 - 2(1/3)
- Multiply: 4/3 + 1 = 3 - 2/3
- Add/Subtract: 7/3 = 7/3
Here we see that 7/3 does indeed equal 7/3.
∴ x = 1/3 is a solution of the equation.
Answer:
8x -16y = 24
Step-by-step explanation:
8x - 16y= -14
A parallel line (i.e. same slope) will be: 8x - 16y= c
We have to find the value of "c".
If it passes through (x, y) = (3, 0) ⇒ 24 - 0 = c,
so equation is 8x -16y = 24
We have the following expression:
(x- (5/2)) ^ 2 = (13/4)
Let's rewrite the given expression:
x ^ 2 - 5x + 25/4 = (13/4)
x ^ 2 - 5x + 25/4 - 13/4 = 0
x ^ 2 - 5x + 12/4 = 0
x ^ 2 - 5x + 3 = 0
Answer:
the original equation given to Sam could have been:
x ^ 2 - 5x + 3 = 0
What is the question? If you are only trying to expand the
expression then the answer would be:
1/4 (5y-3)+ 1/16 (12y+17)
(5y/4) – (3/4) + (12y/16) + (17/16)
1.25y – 0.75 + 0.75y + 1.0625
2y + 0.3125
If you are trying to find for y, then you forgot to equate
it to 0, that is:
2y + 0.3125 = 0
2y = -0.3125
<span>y = 0.15625</span>