You should search it up on Photomath is easier
X²+15x+36<0
at first solve quadratic equation
D=b²-4ac= 225-4*1*36= 81
x=(-b+/-√D)/2a
x=(-15+/-√81)/2= (-15+/-9)/2
x1=(-15-9)/2=-12
x2=(-15+9)/2=-3
we can write x²+15x+36<0 as (x+12)(x+3)<0
(x+12)(x+3)<0 can be 2 cases, because for product to be negative one factor should be negative , and second factor should be positive
1 case) x+12<0, and x+3>0,
x<-12, and x>-3
(-∞, -12) and(-3,∞) gives empty set
or second case) x+12>0 and x+3<0
x>-12 and x<-3
(-12,∞) and (-∞,-3) they are crossing , so (-12, -3) is a solution of this inequality
Given:
Cheshire Road Shelby Road
Plot 1 40 yards 48 yards
Plot 3 x 56 yards
To solve for x, we get the ratio and proportion of both plots.
40/48 = x/56
Do cross multiplication.
40 * 56 = 48 * x
2,240 = 48x
2,240/48 = 48x/48
46.67 = x
Plot 3 of Chesire Road measures 46.67 or 46 2/3 yards. Choice A.
Given:
ABCD is a quadrilateral
AC = 12, DE = 4 and FB = 5
To find:
The area of the polygon.
Solution:
AC bisects the quadrilateral into two triangles.
Area of triangle:
<u>Area of triangle DAC:</u>
Area of triangle DAC = 24 square units.
<u>Area of triangle BAC:</u>
Area of triangle BAC = 30 square units.
Area of polygon = Area of triangle DAC + Area of triangle BAC
= 24 square units + 30 square units
= 54 square units
The area of the polygon is 54 square units.
Answer:
7
Step-by-step explanation:
time=<u>interest</u>
P*R
time=<u>1386*</u>100
6.6 * 3000
<u>time=7</u>
