<span>10!/5! =10*9*8*7*6=30,240;;
If order is not considered, 10C5 =10!/5!5!=252 </span>
Answer: 15e^5x
Step - by - step
y=3e^5x - 2
By the sum rule, the derivative of 3e^5x - 2 with respect to x is d/dx [ 3e^5x ] + d/dx [-2].
d/dx [ 3e^5x ] + d/dx [ -2 ]
Evalute d/dx [ 3e^5x ]
Since 3 is constant with respect to x , the derivative of 3e^5x with respect to x is
3 d/dx [ e^5x ].
3 d/dx [ e^5x ] + d/dx [ -2 ]
Differentiate using the chain rule, which states that d/dx [ f(g(x))] is f' (g(x)) g' (x) where f(x) = e^x and g(x) = 5x.
To apply the Chain Rule, set u as 5x.
3 ( d/du [ e^u] d/dx [5x] ) + d/dx [ -2]
Differentiate using the Exponential rule which states that d/du [ a^u ] is a^u ln(a) where a=e.
3( e^u d/dx[5x] ) + d/dx [ -2 ]
Replace
3(e^5x d/dx [5x] ) + d/dx [ -2 ]
3(e^5x( 5 d/dx [x] )) + d/dx [ -2 ]
Diffentiate using the Power Rule which states that d/dx [x^n] is nx^n-1 where n=1.
3(e^5x(5*1)) + d/dx [-2]
3 ( e^5x * 5 ) + d/dx [-2]
Multiply 5 by 3
15e^5x + d/dx [-2]
Since -2 is constant with respect to x, the derivative of -2 with respect to x is 0.
15e^5x + 0
15e^5x
Answer:
William will make 12 bags of food and each of the bag will contains 2 cans of fruit and 5 cans of vegetables.
Step-by-step explanation:
Given:
Number of fruits cans = 24
Number of veggies cans = 60
William will have to distribute them in equal bags with equal cans of fruits and vegetables respectively.
For this:
We have to find the GCF (greatest common factor) of 24 and 60.
GCF by listing out the factors method.
Factors of 24 :
Factors of 60 : 
So,
The greatest common factor of
and
is
.
The number of bags William will used for equal distribution = 
Now,
We have to distribute the veggies and fruits in equal number of cans to these
bags.
Number of fruits cans used in each bag =
Number of vegetables can used in each bag = 
We can say that:
William will make 12 bags of food and each of the bag will contains 2 cans of fruit and 5 cans of vegetables.