5x+2y2-y3:x=2 and y=4
5(2)+2(4)2-(4)3
10+8(2) -12
10+16-12
26-12
=14
Answer:
3/x^2 or 3x^-2
54x^5
x^-1/3
Step-by-step explanation:
2 6x^3×9x^2=54x^3+2=54x^5
3 x^2/3 × x^-1=x^-1/3
Answer:
Any equation of the line that is different from y=x is the solution to the problem.
Step-by-step explanation:
step 1
Find the slope of the line that passes through the points (1, 1) and (5, 5)
m=(5-1)/(5-1)=1
step 2
Find the equation of the line into slope point form
y-y1=m(x-x1)
we have m=1
(x1,y1)=(1,1)
substitute
y-1=(1)(x-1)
y=x-1+1
y=x
therefore
Any equation of the line that is different from y=x is the solution to the problem.
Answer:
-5
Step-by-step explanation:
12-5=7
The explorer's running velocity will be 1/3 the velocity of the train, 45/3 = 15.
From those issue statement, the train will a chance to be at the begin of the span The point when those voyagers have run 1/3 of the period of the span. In they run of the much end of the bridge, they will bring run 1/3 of the length of the span when those prepare achieves the close to wind. Beginning starting with those 1/3 point, they will make toward those 2/3 perspective At those prepare achieves those span. The prepare will travel the full period of the span in the the long haul it takes those voyagers to run those remaining 1/3. Accordingly the train must travel toward three times the running speed of the voyager. Train = 45 mph, voyagers = 15 mph.
The speed of P&D = 15 mph
