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.
There are 8 possible outcomes for a marble being drawn and numbered.
{1,2,3,4,5,6,7,8}
There are 4 possible outcomes for a card being selected from a standard deck.
{ <span>hearts, diamonds, clubs, spades}
So the number of outcomes in the sample space would be 8 x 4 = 32.
In the event "an even number is drawn", there are only 4 possible outcomes for a marble being drawn, {2,4,6,8}, whereas there are still 4 possible outcomes for a suit. So the number of outcomes in the event is 4 x 4 = 16.
</span><span>In the event "a number more than 2 is drawn and a red card is drawn", there are 6 possible outcomes for the marble being drawn, {3,4,5,6,7,8}, whereas there are only two possible suits for a card being selected as red, {heart, diamond}. So the number of outcomes in this event is 6 x 2 = 12.
In the event </span><span>"a number less than 3 is drawn or a club is not drawn", the number drawn could be 1 or 2 whereas a spade/heart/diamond could be selected. So the number of outcomes is 2 x 3 = 6.</span><span>
</span>
Answer:
This is always ''interesting'' If you see an absolute value, you always need to deal with when it is zero:
(x-4)=0 ===> x=4,
so that now you have to plot 2 functions!
For x<= 4: what's inside the absolute value (x-4) is negative, right?, then let's make it +, by multiplying by -1:
|x-4| = -(x-4)=4-x
Then:
for x<=4, y = -x+4-7 = -x-3
for x=>4, (x-4) is positive, so no changes:
y= x-4-7 = x-11,
Now plot both lines. Pick up some x that are 4 or less, for y = -x-3, and some points that are 4 or greater, for y=x-11
In fact, only two points are necessary to draw a line, right? So if you want to go full speed, choose:
x=4 and x= 3 for y=-x-3
And just x=5 for y=x-11
The reason is that the absolute value is continuous, so x=4 works for both:
x=4===> y=-4-3 = -7
x==4 ====> y = 4-11=-7!
abs() usually have a cusp int he point where it is =0
Step-by-step explanation:
Answer:
i think it would be a 24
Step-by-step explanation:
6×3 = 18
8×3 = 24
