Answer: The correct option is (B) less than 2.
Step-by-step explanation: Given that Y varies directly with X and has a constant rate of change of 6.
We are to find the possible value of X when the value of Y is 11.
Since Y varies directly with X, so we can write
![Y\propto X\\\\\Rightarrow Y=kx~~~~~~~~~~~~~~~(i),~~~~~~~~~\textup{[where 'k' is the constant of variation]}](https://tex.z-dn.net/?f=Y%5Cpropto%20X%5C%5C%5C%5C%5CRightarrow%20Y%3Dkx~~~~~~~~~~~~~~~%28i%29%2C~~~~~~~~~%5Ctextup%7B%5Bwhere%20%27k%27%20is%20the%20constant%20of%20variation%5D%7D)
Since the rate of change is constant and is equal to 6, so we must have
So, from equation (i), we have
Now, when Y = 11, then from equation (ii), we get
Thus, the value of X is less than 2 when Y = 11.
Option (B) is CORRECT.
Answer:
100( 1 - (9/10)^75 )
Step-by-step explanation:
as common ratio is 9/10
Answer:
60ft/s
Step-by-step explanation:
To find the rate
rate = f(3) - f(0)
-------------
3-0
rate = 200-380
---------------
3-0
rate = -180/3 = -60 ft/s
The negative tells us it is falling
The fall falls 60ft/s
I: y=(1/2)x+5
II: y=(-3/2)x-7
substitution:
fancy word for insert the definition of one variable in one equation into the other
-> isolate a variable, luckily y is isolated (even in both equations) already
-> substitute y of II into I (=copy right side of II and replace y in I with it):
(-3/2)x-7=(1/2)x+5
-3x-14=x+10
-3x-24=x
-24=4x
-6=x
-> insert x back into I (or II):
y=(1/2)x+5
=(1/2)*(-6)+5
=-3+5=2
elimination: subtract one equation from the other to eliminate a variable, again y is already isolated->no extra work required
I-II:
y-y=(1/2)x+5-[(-3/2)x-7]
0=(1/2)x+5+(3/2)x+7
0=(4/2)x+12
-12=2x
-6=x
-> insert x back into I (or II):
y=(1/2)x+5
=(1/2)*(-6)+5
=-3+5=2
