You have known the formula Length of base is la division of Area and a haft of the height.
So we have 3
or
and
have fun
Taking the derivative of 7 times secant of x^3:
We take out 7 as a constant focus on secant (x^3)
To take the derivative, we use the chain rule, taking the derivative of the inside, bringing it out, and then the derivative of the original function. For example:
The derivative of x^3 is 3x^2, and the derivative of secant is tan(x) and sec(x).
Knowing this: secant (x^3) becomes tan(x^3) * sec(x^3) * 3x^2. We transform tan(x^3) into sin(x^3)/cos(x^3) since tan(x) = sin(x)/cos(x). Then secant(x^3) becomes 1/cos(x^3) since the secant is the reciprocal of the cosine.
We then multiply everything together to simplify:
sin(x^3) * 3x^2/ cos(x^3) * cos(x^3) becomes
3x^2 * sin(x^3)/(cos(x^3))^2
and multiplying the constant 7 from the beginning:
7 * 3x^2 = 21x^2, so...
our derivative is 21x^2 * sin(x^3)/(cos(x^3))^2
Answers:
c=0.25p
p=4c
c=.25p
c=1/4p
Hope I' not too late -w-'
105 different outfits can wear on an individual day
<em><u>Solution:</u></em>
I have 7 button down shirts, 5 pairs of pants and 3 pairs of shoes that I can wear to work
To find: Number of different outfits can I wear on an individual day
First he has to decide on a pair of pants and he has 5 different choices
For each of those choices he has a choice of 7 different button down shirts, so that gives him 5 x 7 = 35 different pant/shirt combinations
For each of those 35 different pant/shirt combinations, he has 3 pairs of shoes he could select, so altogether he has 35 x 3 = 105 different outfits
In short we can say,
different outfits = 7 x 5 x 3 = 105
So there 105 different outfits can wear on an individual day
