Answer:
11y-2x= -90
Step-by-step explanation:
for the point (1,-8)
x=1 and y= -8
for the equation,
2x -4y= -11
x= -1/m
m= -1/x
x = -1/ (-11/2 )
m= 2/11
so for the equation,
y-y1= m(x- x1)
y- -8=2/11(x-1)
11(y+8)= 2(x-1)
11y +88= 2x-2
11y-2x= -90
Answer:
- Since the question is incomplete, see the figure attached and the explanation below.
Explanation:
Since the figure is missing, I enclose the figure of a square inscribed in a circle.
Since the <em>area of a square</em> is the side length squared, you can determine the side length:

From the side length, you can find the diagonal of the square, which is equal to the diameter of the circle, using the Pythagorean theorem:
- diagonal² = (10cm)² + (10cm)² = 2 × (10cm)²

The area of the circle is π (radius)².
- radius = diameter/2 = diagonal/2

Answer:
see explanation
Step-by-step explanation:
(a)
x² + 2x + 1 = 2x² - 2 ( subtract x² + 2x + 1 from both sides
0 = x² - 2x - 3 ← in standard form
0 = (x - 3)(x + 1) ← in factored form
Equate each factor to zero and solve for x
x + 1 = 0 ⇒ x = - 1
x - 3 = 0 ⇒ x = 3
-----------------------------------
(b)
-
=
( multiply through by 15 to clear the fractions )
5(x + 2) - 2 = 3(x - 2) ← distribute parenthesis on both sides
5x + 10 - 2 = 3x - 6
5x + 8 = 3x - 6 ( subtract 3x from both sides )
2x + 8 = - 6 ( subtract 8 from both sides )
2x = - 14 ( divide both sides by 2 )
x = - 7
--------------------------------------------
(c) Assuming lg means log then using the rules of logarithms
log
⇔ nlogx
log x = log y ⇒ x = y
Given
log(2x + 3) = 2logx
log(2x + 3) = log x² , so
x² = 2x + 3 ( subtract 2x + 3 from both sides )
x² - 2x - 3 = 0
(x - 3)(x + 1) = 0
x = 3 , x = - 1
x > 0 then x = 3
5. To figure this out you must find when y = 0
-16t^2 = -200
t^2 = 12.5
Square root
t = 3.54 C
6.
Axis of symetry = -.5 (always a vertical line down vertex)
Vertex = (-.5, -6.5) (the minimum of the graph)
B
7. To find maximum height you must find the vertex
The maximum height is 549 feet after 5.75 seconds
C
9. Solve for X
x^2 = 1
Square root
X = plus or minus 1
A
They Have 2: <span>Like an ellipse, an hyperbola has two foci and two vertices; unlike an ellipse, the foci in an hyperbola are further from the hyperbola's center than are its vertices: The hyperbola is centered on a point (h, k), which is the "center" of the hyperbola. hope this helps c:</span>