X = smallest odd
x + 2 = second odd
x + 4 = third odd
x + x + 2 + x + 4 = 3
3x + 6 = 3
3x = -3
x = -1
The numbers are -1, 1, 3. Hope this helps!
A function f(x) has solutions if we can find a value to plug in that leads to 0. In other words, there are solutions to f(x) = 0. Another term for "solution" is "root" or "x intercept"
An exponential function may cross the x axis at one point only. Though there are plenty of cases when there are no solutions at all. For instance, in the case of f(x) = (2^x) + 10
The right hand side will never be equal to zero no matter what you plug in for x. The graph will never cross the axis.
To answer your question, yes it is possible to have an exponential equation to have no solutions.
Answer:
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Step-by-step explanation:
Given
---x = 0, in 2012
-- x = 5, in 2017
Required
Select all possible equations
Because there is a reduction in the population, as time increases; the rate must be less than 1.
An exponential function is represented as:
Where
rate > 1 in options (a) and (b) i.e. 1.03
This implies that (a) and (b) cannot be true
For option (c), we have:
Set x = 0
Set x = 5
<em>This is true because the calculated values of f(0) and f(5) correspond to the given values</em>
For option (d), we have:
Set x = 0
<em>This is false because the calculated value of f(0) does not correspond to the given value</em>
For option (e), we have:
Set x = 0
undefined
<em>This is false because the f(x) is not undefined at x = 0</em>
For option (f), we have:
Set x = 0
<em>This is false because the calculated value of f(0) does not correspond to the given value</em>
<em>From the computations above, only (c) </em><em> is true</em>
Answer:
sum
Step-by-step explanation:
The cross product of the normal vectors of two planes result in a vector parallel to the line of intersection of the two planes.
Corresponding normal vectors of the planes are
<5,-1,-6> and <1,1,1>
We calculate the cross product as a determinant of (i,j,k) and the normal products
i j k
5 -1 -6
1 1 1
=(-1*1-(-6)*1)i -(5*1-(-6)1)j+(5*1-(-1*1))k
=5i-11j+6k
=<5,-11,6>
Check orthogonality with normal vectors using scalar products
(should equal zero if orthogonal)
<5,-11,6>.<5,-1,-6>=25+11-36=0
<5,-11,6>.<1,1,1>=5-11+6=0
Therefore <5,-11,6> is a vector parallel to the line of intersection of the two given planes.
