Given that JKLM is a rhombus and the length of diagonal KM=10 na d JL=24, the perimeter will be found as follows;
the length of one side of the rhombus will be given by Pythagorean theorem, the reason being at the point the diagonals intersect, they form a perpendicular angles;
thus
c^2=a^2+b^2
hence;
c^2=5^2+12^2
c^2=144+25
c^2=169
thus;
c=sqrt169
c=13 units;
thus the perimeter of the rhombus will be:
P=L+L+L+L
P=13+13+13+13
P=52 units
Left side: 63, 72, 16 right side: 36, 4, 35
Answer:
Somewhere around 27k
Step-by-step explanation:
Answer:
Step-by-step explanation:
Our function is : 3x²-3x-1
- 3x²-3x+2=2
- 3x²-3x-1+3=2
- 3x²-3x-1= -1
- the number that gives us -1 is 1 (from the graph)
so x is 1
- 3x²-3x-1= x+1 ⇒3x²-4x-2 = 0
- Δ(the dicriminant)= (-4)²-4*3*(-2)= 16+24 =40
- two solution x and y
- x= (4-√40)/6 and y= (4+√40)/6
- x= -0.38 and y = 1.7
- so the estimations are -0.38 and 1.7
Answer:
A number line going from 27 to 37, open circle at 32, everything to the right is shaded
Step-by-step explanation:
-1/2p < -16
multiply each side by -2 (don't forget to switch inequality symbol)
p > 32
