By factorising I got these solutions
X = 0
X = 2
X = - 2
X = 3
<span>Starting position:
Let B start at origin(O) & A is at (-150 ,0) ie 150 km west of the origin . At 3pm , t= 0
A is changing its position at ( 150-35t) km/h
& B is changing its position at 25t km/h
Therefore
AB^2 = OA^2 + OB^2 /// Pythagoras Theorem
= (150-35t)^2 + ( 25t)^2
= 22 500 - 10 500 t + 1225t^2 +625t^2
= 22 500 - 10 500t + 1850t^2
AB = ( 22 500 - 10 500t + 1850t^2)^1/2
d (AB) /dt = 1/2 * ( 22 500 - 10 500t + 1850t^2)^-1/2 * ( - 10 500 + 3700t)
= ( - 5250 + 1850t) / ( 22 500 - 10 500t + 1850t^2)^1/2
At 3pm , t=0
At 7pm , t= 4
So d (AB) /dt = ( -5250 + 1850*4) / ( 22500 - 10500*4 + 1850*4^2)^1/2
= 2150 / 100.5
= 21.4 km/hr
</span>The distance between the ships changing at 7 P.M. is with a speed of 21.4 km/hr.
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
Answer:
A
Step-by-step explanation:
You could raise 1 ^ 1000000 and the answer would still be 1.
Ogetchi is wrong.
Bernardo is right
a1 = 1
r = 8
Each term is 8 times the number before it. 8*8 = 64 ; 8*64 = 512
A
Answer:
Step-by-step explanation:
for question 5 )
we know that x-5 = 6 because I'm pretty sure outside of the question we are told that the rectangles are similar and therefore, we can say the above
x-5 = 6
add 5 to both side
x = 11
for question 6)
30 is 
of 36, so use that on the remaining given side
* 15 = 12.5
then thats what the other side of the bigger shape is
12.5 = x + 6
minus 6 from both sides
6.5 = x
by the way, notice that they have used a variable "x" for both 5 & 6 which is named the same but has a different value in each problem. X is the name of the variable for each problem, but doesn't have any meaning once outside of each of the problems. It's just confusing to use the same named variable at times. something like xy and xz , might have been better names. anyway :)
