Answer:
X= -12 +6√10
X= 6.973
Step-by-step explanation:
Volume of the box= (36-2x)(24-2x)x
Volume of the box
=( 864 -96x -4x²)x
But 864 -96x -4x²
=216 - 24x-x²
Solving for x quadratically
X= (24+12√10)/-2
X= -12 -6√10
X= -30.97
Or
X= (24-12√10)/-2
X= -12 +6√10
X= 6.973
X will definitely be a positive number
So X= -12 +6√10
X= 6.973
A method that always works is to find the slope of the given line, then find the negative reciprocal of that. Your result will be the slope of the perpendicular line. Using this slope and the given point, fill in the parameters of the point-slope form of the equation of a line.
For m = slope of given line and (h, k) = given point, the perpendicular line will be
y = (-1/m)(x -h) +k
Often, this equation can be simplified to another appropriate form, such as slope-intercept form (y = mx+b) or standard form (ax+by=c).
_____
The slope of a given line can be found by solving its equation for y. The slope is the coefficient of x in that solution. If the given line is characterized by two points, (x1, y1) and (x2, y2), then its slope is m = (y2-y1)/(x2-x1).
In the unusual case where the given line is vertical (x=<some constant>), the slope of the perpendicular line is zero, and the line you want becomes y=k.
To find the unit rate, you would do 2.99 / 11.3 and get about 26 cents. So the unit rate would be 26 over 1 ( 26/1 ).
Hope I helped!
Marsh
Answer:
Step-by-step explanation:
9) PQR Is an isosceles triangle
=> ∠PRQ = (180° - x)/2
PRS is an isosceles right triangle
=> ∠PRS = 45°
Have: ∠PRS + ∠PRQ = 115°
=> 
=> 180 - x = (115 - 45).2 = 140
<=> x = 180 - 140 = 40
10) ABD is an isosceles right triangle => ∠ABD = 45°
BCD is an equilateral triangle => ∠CBD = 60°
have: x = ∠ABD + ∠CBD = 45° + 60° = 105°
11) have: x = y (2)
PQT is an isosceles triangle => ∠PQT = 180 - 70.2 = 40
QTS is an isosceles triangle => ∠TQS = 180 -2x
QRS is an isosceles triangle => ∠RSQ = y
have: 40 + 180 - 2x + y = 180 => 2x - y = 40 (1)
(1)(2) => 
=> x + y = 80
12) EFJ Is an equilateral triangle => ∠FJE = 60
∠FJE is the outer angle of the triangle FHJ but FHJ is an isosceles triangle
=> 60 = 2.∠JHF => ∠JHF = 30°
∠JHF is the outer angle of the triangle FHG
=> 30° = 2x
<=> x = 15°
