Given equation : 
.
Strategy 1: We can cross mutiply both sides remove fraction form.
On cross multiplication, we get
x * 7 = 3 * 42
7x = 126.
Dividing both sides by 7, we get
<h3>
x = 18.</h3>
Strategy 2: We can find least common denominator(lcd) of both sides and multiplying both sides by that lcd to get rid denominators from both sides.
LCD of 42 and 7 is 42.
Therefore, multiplying both sides by 42, we get
x = 6 * 3
<h3>x = 18.</h3>
0.00100037227 will be your correct answer.
Answer:
Step-by-step explanation:
11) strategy: since they tell us. indirectly, that the length of JK is the same as LM then we can set those two equal, solve for X and then.. when we have X we can figure out the lenght of JK and LM and then just divide that by 2 go get PK ( :0 Player Killer ??? no , not that PK)
solve JK and LM
3x + 23 =9x-19
42 = 6x
7 = x
now that we know x = 7 plug it into either equation to come up with the length of JK or LM . I'll pick JK just b/c it was 1st
3(7) + 23 = JK
21+23 = JK
44 = JK
now take half of 1/2*JK= 22 that is PK ( are you sure that 's not player killer?)
PK = 22
12) strategy: set the two arcs BG and GC equal and solve for X, then plug x into either equation and the multiply the answer by 2 to find arc AB
9x-20 = 5x + 28
4x = 48
X = 12
9(12) -20 = BG
88 = BG
2*88 + AB
176 = AB
13) done
14) strategy: find the angle at L, and that will also be the arc of MK
<em>copy and past the below</em> helpful trig functions into your computer
Use SOH CAH TOA to recall how the trig functions fit on a triangle
SOH: Sin(Ф)= Opp / Hyp
CAH: Cos(Ф)= Adj / Hyp
TOA: Tan(Ф) = Opp / Adj
<em>copy and past the above</em> :
use which ever trig function you want , we have all the sides of the triangle, I'll use CAH
Cos(Ф) = 9/15
Ф = arcCos(9/15)
Ф = 53.13010°
arc MK = 53.13010°
15) strategy: arc JK is just 2 times MK
2*MK = 106.26020°
arc MK = 106.26020°
16) find arc JPK strategy: JPK is just the remaining part of a full circle of 360 - MK = 253.7397°
arc JPK = 253.7397°
Answer:
144
Step-by-step explanation:
im not confident in this answer but whatever
