Answer:
b
Step-by-step explanation:
Step-by-step explanation:
The error carolyn has made is that she hasn't subtracted properly.
2x + 3y - (2x +6y) = 2x - 2x + 3y - 6y = -3y.
Carolyn probably didn't use parentheses...
To do the thing in substitution, we single out x or y in any equation. In this case, i'm gonna single out the x in the 2nd eqn. In other words, i'll make x the subject of the 2nd eqn.
2x + 6y = 10
2x = 10 - 6y
x = 5 - 3y
Now we simply plug in 5 - 3y wherever x apperas in the 1st equation.
2 (5-3y) + 3y = 8
10 - 6y + 3y = 8
10 - 3y = 8
-3y = -2
y = 2/3
Then x = 5 - 3*2/3 = 3
x=3
y=2/3
Hope it helps:)
Answer:
f'(x) > 0 on
and f'(x)<0 on
Step-by-step explanation:
1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:

To find its decreasing interval :

2) Then let's find the critical point of this function:
![f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[6-2^{2x}]=\frac{\mathrm{d} }{\mathrm{d}x}[6]-\frac{\mathrm{d}}{\mathrm{d}x}[2^{2x}]=0-[ln(2)*2^{2x}*\frac{\mathrm{d}}{\mathrm{d}x}[2x]=-ln(2)*2^{2x}*2=-ln2*2^{2x+1\Rightarrow }f'(x)=-ln(2)*2^{2x}*2\\-ln(2)*2^{2x+1}=-2x^{2x}(ln(x)+1)=0](https://tex.z-dn.net/?f=f%27%28x%29%3D%5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cmathrm%7Bd%7D%20x%7D%5B6-2%5E%7B2x%7D%5D%3D%5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cmathrm%7Bd%7Dx%7D%5B6%5D-%5Cfrac%7B%5Cmathrm%7Bd%7D%7D%7B%5Cmathrm%7Bd%7Dx%7D%5B2%5E%7B2x%7D%5D%3D0-%5Bln%282%29%2A2%5E%7B2x%7D%2A%5Cfrac%7B%5Cmathrm%7Bd%7D%7D%7B%5Cmathrm%7Bd%7Dx%7D%5B2x%5D%3D-ln%282%29%2A2%5E%7B2x%7D%2A2%3D-ln2%2A2%5E%7B2x%2B1%5CRightarrow%20%7Df%27%28x%29%3D-ln%282%29%2A2%5E%7B2x%7D%2A2%5C%5C-ln%282%29%2A2%5E%7B2x%2B1%7D%3D-2x%5E%7B2x%7D%28ln%28x%29%2B1%29%3D0)
2.2 Solving for x this equation, this will lead us to one critical point since x' is not defined for Real set, and x''
≈0.37 for e≈2.72

3) Finally, check it out the critical point, i.e. f'(x) >0 and below f'(x)<0.
Given the table below which shows the result of a survey that asked 2,881 people whether they are involved in any type of charity work.
![\begin{tabular} {|c|c|c|c|c|c|} &Frequently&Occassionally&Not at all&Total\\[1ex] Male&227&454&798&1,479\\ Female &205&450&747&1,402\\ Total&432&904&1,545&2,881 \end{tabular}](https://tex.z-dn.net/?f=%5Cbegin%7Btabular%7D%0A%7B%7Cc%7Cc%7Cc%7Cc%7Cc%7Cc%7C%7D%0A%20%26Frequently%26Occassionally%26Not%20at%20all%26Total%5C%5C%5B1ex%5D%0AMale%26227%26454%26798%261%2C479%5C%5C%0AFemale%20%26205%26450%26747%261%2C402%5C%5C%0ATotal%26432%26904%261%2C545%262%2C881%0A%5Cend%7Btabular%7D)
Part A:
If a person is chosen at random, the probability that the person is frequently or occassinally involved in charity work is given by

Part B:
If a person is chosen at random, the probability that the person is female or not involved in charity work at all is given by

Part C:
If a person is chosen at random, the probability that the person is male or frequently involved in charity work is given by

Part D:
If a person is chosen at random, the probability that the person is female or not frequently involved in charity work is given by

Part E:
The events "being female" and "being frequently involved in charity work" are not mutually exclusive because being a female does not prevent a person from being frequently involved in charity work.
Indeed from the table, there are 205 females who are frequently involved in charity work.
Therefore, the answer to the question is "No, because 205 females are frequently involved charity work".
Answer:
x=-7
Step-by-step explanation: