Answer:
118
Step-by-step explanation:
Answer:
How to solve your problem
−
7
2
−
2
2
+
3
−
2
+
5
3
−
2
-7y^{2}-2y^{2}+y^{3}y-2y+5y^{3}-2y
−7y2−2y2+y3y−2y+5y3−2y
Simplify
1
Combine exponents
−
7
2
−
2
2
+
3
−
2
+
5
3
−
2
-7y^{2}-2y^{2}+{\color{#c92786}{y^{3}y}}-2y+5y^{3}-2y
−7y2−2y2+y3y−2y+5y3−2y
−
7
2
−
2
2
+
4
−
2
+
5
3
−
2
-7y^{2}-2y^{2}+{\color{#c92786}{y^{4}}}-2y+5y^{3}-2y
−7y2−2y2+y4−2y+5y3−2y
2
Combine like terms
−
7
2
−
2
2
+
4
−
2
+
5
3
−
2
{\color{#c92786}{-7y^{2}}}{\color{#c92786}{-2y^{2}}}+y^{4}-2y+5y^{3}-2y
−7y2−2y2+y4−2y+5y3−2y
−
9
2
+
4
−
2
+
5
3
−
2
{\color{#c92786}{-9y^{2}}}+y^{4}-2y+5y^{3}-2y
−9y2+y4−2y+5y3−2y
3
Combine like terms
−
9
2
+
4
−
2
+
5
3
−
2
-9y^{2}+y^{4}{\color{#c92786}{-2y}}+5y^{3}{\color{#c92786}{-2y}}
−9y2+y4−2y+5y3−2y
−
9
2
+
4
−
4
+
5
3
-9y^{2}+y^{4}{\color{#c92786}{-4y}}+5y^{3}
−9y2+y4−4y+5y3
4
Rearrange terms
−
9
2
+
4
−
4
+
5
3
{\color{#c92786}{-9y^{2}+y^{4}-4y+5y^{3}}}
−9y2+y4−4y+5y3
4
+
5
3
−
9
2
−
4
{\color{#c92786}{y^{4}+5y^{3}-9y^{2}-4y}}
y4+5y3−9y2−4y
Solution
4
+
5
3
−
9
2
−
4
Answer: G(x) = 3|x + 2| - 8
<u>Step-by-step explanation:</u>
"a" represents the vertical stretch <em>(or shrink)</em>
"h" represents the x-coordinate of the vertex
- x-axis goes left and right
"k" represents the y-coordinate of the vertex
F(x) = 3|x - 2| - 5
G(x) = 3(x - 2 + 4| - 5 - 3
= 3|x + 2| - 8
Answer: h = 190.
Step-by-step explanation: You have to use the reverse so, the opposite of subtraction is addition, so 156 + 34 = 190. 190 which is the "h" 190 - 156 = 34
Splitting up [0, 3] into 
equally-spaced subintervals of length 
gives the partition
![\left[0, \dfrac3n\right] \cup \left[\dfrac3n, \dfrac6n\right] \cup \left[\dfrac6n, \dfrac9n\right] \cup \cdots \cup \left[\dfrac{3(n-1)}n, 3\right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%20%5Cdfrac3n%5Cright%5D%20%5Ccup%20%5Cleft%5B%5Cdfrac3n%2C%20%5Cdfrac6n%5Cright%5D%20%5Ccup%20%5Cleft%5B%5Cdfrac6n%2C%20%5Cdfrac9n%5Cright%5D%20%5Ccup%20%5Ccdots%20%5Ccup%20%5Cleft%5B%5Cdfrac%7B3%28n-1%29%7Dn%2C%203%5Cright%5D)
where the right endpoint of the 
-th subinterval is given by the sequence
for 
.
Then the definite integral is given by the infinite Riemann sum
