My brain capacity can not handle this one right now? Anyone good with geometry word problems? It’s number 12 if anyone is able t
o help out you are a blessing
1 answer:
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Derivative of the denominator:
Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.
and then split the fraction,
Based on our previous test, we know that a simple substitution will work for the first integral:
So the first integral changes,
integrating to a log,
Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.
So we have this going on,
Let's factor the 5 out of the intergral,
and the 4 from the denominator,
Bringing all that stuff together as a single square,
Making the substitution:
giving us,
simplying a lil bit,
and hopefully from this point you recognize your arctangent integral,
undo your substitution as a final step,
and include a constant of integration,
Hope that helps!
Lemme know if any steps were too confusing.
The number of possibilities for constructing the 4 characters long password with specified conditions is given by: Option C: 703,040
<h3>What is the rule of product in combinatorics?</h3>If a work A can be done in p ways, and another work B can be done in q ways, then both A and B can be done in ways.
Remember that this count doesn't differentiate between order of doing A first or B first then doing other work after the first work.
Thus, doing A then B is considered same as doing B then A
We're specified that:
- The password needs to be 4 characters long
- It must have 3 letters and 1 of 10 special characters.
- Repetition is allowed.
So, each of 3 characters get 26 ways of being 1 letter. (assuming no difference is there between upper case letter and lower case letter).
And that 1 remaining character get 10 ways of being a special character.
So, by product rule, this choice (without ordering) can be done in:
ways.
Now, the password may look like one of those:
- L, L, L, S
- L, L, S, L
- L, S, L, L
- S, L, L, L
where S shows presence of special character and L shows presence of letter.
Those 175760 ways are available for each of those four ways.
Thus, resultant number of ways this can be done is:
Thus, the number of possibilities for constructing the 4 characters long password with specified conditions is given by: Option C: 703,040
Learn more about rule of product here: