Remember x is x and f(x) is same thing as y.
f(5) = 4
It is same thing as saying, when x = 5, y aka f(x) should be 4. Aka the point (5,4)
Let’s look at D, if x =5 then f(x)=5? No, it should equal 4. So this is wrong. It is also not linear. So NOT D
Lets look at C, if x =5 then
1/5(5)+3 =
5/5 + 3 =
1+3 =
4
So, f(x) aka y = 4. Is this correct? Yes when x =5, y indeed should equal 4, but it is not linear. So NOT C
Lets look at B, if x=5 then,
2^5 -28 =
32 - 28 =
4
So, f(x) aka y = 4. Is this correct? Yes when x =5, y indeed should equal 4, AND it is linear SO THE ANSWER IS B
Check A too, it gets y=14 so thats wrong as it is not 4.
So the formula is
7*(1/3+1/5) = x
First find the common denominator of the fractions. In this case the common denominator of 3 and 5 is 15 so we must convert them to 15ths. To do that we divide 15 by the denominator and take that answer and multiply the numerator by it.
So for 1/3 we take 15/3(denominator) = 5 and multiply the numerator (1) by it
5 * 1 = 5 so 1/3 converts to 5/15
Do the same for 1/5 we take 15/5 = 3 and multiply the numerator (4) by it
3*4 = 12 so 4/5 converts to 12/15
Now we add 5/15 and 12/15 = 17/15 so we have our formula now to
7*17/15
We make seven a fraction 7/1 and multiply across 7*17 = 119 and 15 * 1 = 15
We have 119/15 which when we divide and get 7 14/15 as your answer
As Mifc8 said, one can easily do 1.7*1080=1836 feet.
Let the point be P, and the masses be located at P1 and P2.
The centre of mass is such that the moments of masses m1 and m2 exert an equal moment at the point P,
and the distance P1P2=d
namely,
m1(mPP1)=m2(mPP2)
The distance of the centre of mass from m1 is therefore
d1=d*m2/(m1+m2)
Similarly, the distance of the centre of mass from m2 is
d2=d*m1/(m1+m2)
