3y + 5x = -15
3y = -5x - 15 (Subtract 5x from both sides)
y = -5/3x - 5 (Divide everything by 3)
Answer:
Both the boats will closet together at 2:21:36 pm.
Step-by-step explanation:
Given that - At 2 pm boat 1 leaves dock and heads south and boat 2 heads east towards the dock. Assume the dock is at origin (0,0).
Speed of boat 1 is 20 km/h so the position of boat 1 at any time (0,-20t),
Formula : d=v*t
at 2 pm boat 2 was 15 km due west of the dock because it took the boat 1 hour to reach there at 15 km/h, so the position of boat 2 at that time was (-15,0)
the position of boat 2 is changing towards east, so the position of boat 2 at any time (-15+15t,0)
Formula : D=
⇒
Now let
∵
⇒ t= 450/1250
⇒ t= .36 hours
⇒ = 21 min 36 sec
Since F"(t)=0,
∴ This time gives us a minimum.
Thus, The two boats will closet together at 2:21:36 pm.
83 hope it helps get a good grade
Answer:
Step-by-step explanation:
To move the graph 6 units to the right, subtract 6 in the absolute value.
y=|x+5-6|
To move it up, add 2 outside the absolute value.
y=|x+5-6|+2
Now remove the absolute values since x>1.
y=x+5-6+2
y=x+1
I'll start 18 and 22 for you, and you should then be able to do the rest on your own!
For 18, what we literally do is apply the distance formula for all the points and add them up. For B to C, we get the distance between them to be
sqrt((x1-x2)^2+(y1-y2)^2)=sqrt((0-4)^2+(3-(-1))^2)=sqrt((-4)^2+4^2)=sqrt(16+16)=sqrt(32). Repeat the process for C to E, E and F, and F to B then add the results up to get your answer!
For 22, since the area of a rectangle is length*width (we know given the right angles and that the opposite sides are equal in how long they are), we can multiply 2 perpendicular lines, for example, BC and CE to get sqrt(32)*sqrt(8)=16 as the area