14= A.39°
You solved this one so I'm not going to show the work for it.
15=B,C,D
a.-30(-3x+2)=90x-60- incorrect
b.30(-3x+2)=-90x+60
c.10(-9x+6)=-90x+60
d.-10(-9x-6)=90x+60
e.-10(9x+6)=-90x-60- incorrect
16=D
-18x-12-(-13x+17)
-18x-12+13x+17
-5x+5
17=C
9x+3-5x-7
4x-4
18=C
You solved this one...
19=B
8.4x+2.9+(-3.7x+5)
8.4x+2.9-3.7x+5
4.7x+7.9
20=C
21=D
D is the only one that ends with +32 and not -32, which would be incorrect.
22=Cant see >.<
23=A
4x+12+15-3x
x+27
24=B
2x+3
Answer:
1. L1 = 
2. O2 = 
Step-by-step explanation:
Given: x = 
1. To make L1 the subject of formula;
x =
cross multiply to have;
L2 - L1 = xL1(O2 - O1)
collect like terms,
L2 = xL1(O2 - O1) + L1
factorize the right hand side;
L2 = L1[x(O2 - O1)]
L1 =
2. To make O2 the subject of formula;
x =
cross multiply to have;
L2 - L1 = xL1(O2 - O1)
open the bracket to have;
L2 - L1 = xL1O2 - xL1O1
⇒ xL1O2 = L2 - L1 + xL1O1
O2 = 
=
Step-by-step explanation:
according to pythagoras theorem
Hypotenuse2 = Perpendicular2 + Base2
c2 = a2 + b2
therefore,
x2 = (4)2 + (4)2
x2 = 16 + 16
x2 = 32
x = √32
hence, x = 4√2
<span>120 - 36 is 84 just as 12(10 - 3) = 12 x 7 = 84.</span>
Answer:
DB = 6
Step-by-step explanation:
See the attached figure to the question.
Since O is the center of the given circle, so AB and CD are diameters of the circle.
Hence, AO = CO = OD = OB = radius of the circle = 6
In Δ OAC, we have AO = OC. Hence, ∠ A = ∠C = 60° {Given that ∠OAC = 60°}
So, ∠ AOC will automatically become 60°. {As the sum of all the angles of a triangle is 180°}
So, Δ AOC is equilateral triangle.
Now, ∠ AOC = ∠ BOD = 60° {Since, they are vertically opposite angles}
Now, in Δ BOD, we have DO = OB. Hence, ∠ D = ∠ B = 60° {As the sum of all the angles of a triangle is 180°}
So, Δ BOD is also an equilateral triangle.
So, DB = BO = DO = 6. (Answer)
