Answer:
The Taylor series of f(x) around the point a, can be written as:
Here we have:
f(x) = 4*cos(x)
a = 7*pi
then, let's calculate each part:
f(a) = 4*cos(7*pi) = -4
df/dx = -4*sin(x)
(df/dx)(a) = -4*sin(7*pi) = 0
(d^2f)/(dx^2) = -4*cos(x)
(d^2f)/(dx^2)(a) = -4*cos(7*pi) = 4
Here we already can see two things:
the odd derivatives will have a sin(x) function that is zero when evaluated in x = 7*pi, and we also can see that the sign will alternate between consecutive terms.
so we only will work with the even powers of the series:
f(x) = -4 + (1/2!)*4*(x - 7*pi)^2 - (1/4!)*4*(x - 7*pi)^4 + ....
So we can write it as:
f(x) = ∑fₙ
Such that the n-th term can written as:
It is the same distance apart. .1875-.125 is .0625
1/4 is .25-.1875 is .0625
Answer:
1)-
How to solve your question
Your question is
4(4−72)−9(5+2)
4(4y-7y^{2})-9(5y+2)4(4y−7y2)−9(5y+2)
Simplify
1
Rearrange terms
4(4−72)−9(5+2)
4({\color{#c92786}{4y-7y^{2}}})-9(5y+2)4(4y−7y2)−9(5y+2)
4(−72+4)−9(5+2)
4({\color{#c92786}{-7y^{2}+4y}})-9(5y+2)4(−7y2+4y)−9(5y+2)
2
Distribute
4(−72+4)−9(5+2)
{\color{#c92786}{4(-7y^{2}+4y)}}-9(5y+2)4(−7y2+4y)−9(5y+2)
−282+16−9(5+2)
{\color{#c92786}{-28y^{2}+16y}}-9(5y+2)−28y2+16y−9(5y+2)
3
Distribute
−282+16−9(5+2)
-28y^{2}+16y{\color{#c92786}{-9(5y+2)}}−28y2+16y−9(5y+2)
−282+16−45−18
-28y^{2}+16y{\color{#c92786}{-45y-18}}−28y2+16y−45y−18
4
Combine like terms
2)
−17y+17z+24
See steps
Step by Step Solution:

STEP1:Equation at the end of step 1
((24 - 4 • (5y - 6z)) + 3y) - 7z
STEP2:
Final result :
-17y + 17z + 24
−282+16−45−18
-28y^{2}+{\color{#c92786}{16y}}{\color{#c92786}{-45y}}-18−28y2+16y−45y−18
−282−29−18
-28y^{2}{\color{#c92786}{-29y}}-18−28y2−29y−18
Solution
−282−29−18
7/10 because you add seven and three which equals ten, and seven of them are dimes.
Answer:
clean
Step-by-step explanation:
