We are usually concerned with one reaction. That is, the production of one specific set of products from a specific set of reactants.
The number of values of c/d would be the number of possible ways that a and b could recombine to form different pairs of products c and d. (You might get different reactions at different temperatures, for example. Or, you might get different pars of ions.)
Usually, the number of values of c/d is one (1). (Of course, if you simply swap what you're calling "c" and "d", then you double that number, whatever it is.)
Answer:
The correct option is;
The ranges of f(x) and h(x) are similar and different from the ranges of g(x)
Step-by-step explanation:
Here we have the functions given as follows;
f(x) = -6/11 (11/2)ˣ
g(x) = 6/11 (11/2)⁻ˣ
h(x) = -6/11 (11/2)⁻ˣ
from the above equations, it can be seen that f(x) and h(x) are always negative while g(x) is always positive
f(x) and h(x) are symmetric about the y axis while g(x) and h(x) are symmetric about the x axis
Also h(x) is the inverse of f(x) hence the ranges of f(x) and g(x) are similar and are different from the ranges of g(x).
Answer:
Step-by-step explanation:
The problem relates to filling 8 vacant positions by either 0 or 1
each position can be filled by 2 ways so no of permutation
= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
= 256
b )
Probability of opening of lock in first arbitrary attempt
= 1 / 256
c ) If first fails , there are remaining 255 permutations , so
probability of opening the lock in second arbitrary attempt
= 1 / 255 .
Simple :)
Area of shaded part = are of 1/4 circle - area of both triangles.
Are of circle = pie r^2 so 100x3.14 = 314 cm2.
Area of triangle AOB= area of triangleDOE = bh/2= 5x10/2= 25 each
However, the traingles share a common area which is quad DOB(I)
Lets take traingle AOE, whose are is bh/2=10x10/2=50cm2.
50-area of triangle A(I)E= 50-(
Answer:
Option C. The time in seconds that passed before the printer started printing pages
see the explanation
Step-by-step explanation:
Let
y ---->the number of pages printed.
x ---> the time (in seconds) since she sent a print job to the printer
we know that
The x-intercept is the value of x when the value of y is equal to zero
In the context of the problem
The x-intercept is the time in seconds that passed before the printer started printing pages (the number of pages printed is equal to zero)
