if ABC is a isosceles triangle with sides(2a+b)cm, (6a+2b+1)cm and a base of 4a cm, if it's perimeter is 28 find sides
1 answer:
Answer:
a=1.5, b= 3
Step-by-step explanation:
sum of all sides will give us perimeter. Therefore
2a+b+6a+2b+1+4a=28
12a+3b+1=28
12a+3b=27....eqn1
since t two sides of this isosceles triangle are equal:
2a+b=4a
2a=b
substituting in the first equation whenever we encounter a we will have:
6b+3b=27
9b=27
b=3
therefore
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Answer:
(a) k'(0) = f'(0)g(0) + f(0)g'(0)
(b) m'(5) =
Step-by-step explanation:
(a) Since k(x) is a function of two functions f(x) and g(x) [ k(x)=f(x)g(x) ], so for differentiating k(x) we need to use <u>product rule</u>,i.e.,
this will give <em>k'(x)=f'(x)g(x) + f(x)g'(x)</em>
on substituting the value x=0, we will get the value of k'(0)
{for expressing the value in terms of numbers first we need to know the value of f(0), g(0), f'(0) and g'(0) in terms of numbers}{If f(0)=0 and g(0)=0, and f'(0) and g'(0) exists then k'(0)=0}
(b) m(x) is a function of two functions f(x) and g(x) [ ]. Since m(x) has a function g(x) in the denominator so we need to use <u>division rule</u> to differentiate m(x). Division rule is as follows :
this will give <em></em>
on substituting the value x=5, we will get the value of m'(5).
{for expressing the value in terms of numbers first we need to know the value of f(5), g(5), f'(5) and g'(5) in terms of numbers}
{NOTE : in m(x), g(x) ≠ 0 for all x in domain to make m(x) defined and even m'(x) }
{ NOTE : }