The coefficient of y∧15×∧2 in expansion of (y∧3+x)∧7 is (15×2)+(3×7)= 51
All real numbers except -4 and 9.
The domain of an expression is the range of all values for which the expression is defined. Looking at the expression, it will legal everywhere except when a^2 - 5a - 36 is equal to 0. Since that's a quadratic equation, we can use the quadratic formula to find where its value is 0, giving -4 and 9.
Therefore the domain of the expression is, All real numbers except -4 and 9.
(15,-25)(-12,11)
slope = (11 - (-25) / (-12 - 15) = - 36 / 27 which reduces to - 4/3
y = mx + b
slope(m) = -4/3
use either of ur points (15,-25)...x = 15 and y = -25
now we sub and find b, the y int
-25 = -4/3(15) + b
-25 = - 20 + b
-25 + 20 = b
-5 = b
so ur equation is : y = -4/3x - 5 <==
The given inequality holds for the open interval (2.97,3.03)
It is given that
f(x)=6x+7
cL=25
c=3
ε=0.18
We have,
|f(x)−L| = |6x+7−25|
= |6x−18|
= |6(x−3)|
= 6|x−3|
Now,
6|x−3| <0.18 then |x−3|<0.03 ----->−0.03<x-3<0.03---->2.97<x<3.03
the given inequality holds for the open interval (2.97,3.03)
For more information on inequality click on the link below:
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Although part of your question is missing, you might be referring to this full question: For the given function f(x) and values of L,c, and ϵ0, find the largest open interval about c on which the inequality |f(x)−L|<ϵ holds. Then determine the largest value for δ>0 such that 0<|x−c|<δ→|f(x)−|<ϵ.
f(x)=6x+7,L=25,c=3,ϵ=0.18
.
10 because it is teh number but with the opposite sign opposite of a posiive is a negative
