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
1. . D . Identity , when you odd both properties it will always to be true on both sides
Answer:
Step-by-step explanation:
Given that prices for a pair of shoes lie in the interval
[80,180] dollars.
Delivery fee 20% of price.
i.e. delivery fee will be in the interval [4, 9]
(1/20th of price)
Total cost= price of shoedelivery cost
Hence f(c) = c+c/20 = 21c/20
The domain of this function would be c lying between 80 to 180
So domain =[80,180]
---------------------------------
Amount to be repaid = 42 dollars
Once he received this amount, the price would be
105+42 =147
But since price range is only [21*80/20, 21*180/20]
=[84, 189]
Since now Albert has 147 dollars, he can afford is
[80,147]
Answer:
y=11
x=-4
Step-by-step explanation:
y =-x+7
2x-5y=-7
insert the first equation into the second like so...
2x+5x-35=-7
Now rearrange...
2x+5x=-7+35
Now solve...
7x=-28
x=-4
Now incert x in any equation...
y=4+7
y=11
