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erastovalidia [21]
2 years ago
5

Help plsssssssssssssssssssss

Mathematics
1 answer:
noname [10]2 years ago
7 0

Answer:

Question 10:

Answer: b.

{ \tt{f(x) = 5 {x}^{2}  + 9x - 4}} \\ { \tt{g(x) =  -  {8x}^{2} - 3x - 4 }}

(f + g)x, add f(x) and g(x):

{ \tt{(f + g)x = (5 - 8) {x}^{2}  + (9 - 3)x - 4 - 4}} \\ { \tt{(f + g)x =  - 3 {x}^{2}  + 6x - 8}}

Question 11:

Answer: a.

In relation with solution of question 10, same procedure:

{ \tt{(f - g)x =  - 3 {x}^{3} +  (1 - 2) {x}^{2}  + ( - 3 - 4)x + 9 - ( - 9)}} \\ { \tt{(f - g)x =  - 3 {x}^{3} -  {x}^{2}  - 7x + 18 }}

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30

Step-by-step explanation:

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Integration of ∫(cos3x+3sinx)dx ​
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Answer:

\boxed{\pink{\tt I =  \dfrac{1}{3}sin(3x)  - 3cos(x) + C}}

Step-by-step explanation:

We need to integrate the given expression. Let I be the answer .

\implies\displaystyle\sf I = \int (cos(3x) + 3sin(x) )dx \\\\\implies\displaystyle I = \int cos(3x) + \int sin(x)\  dx

  • Let u = 3x , then du = 3dx . Henceforth 1/3 du = dx .
  • Now , Rewrite using du and u .

\implies\displaystyle\sf I = \int cos\ u \dfrac{1}{3}du + \int 3sin \ x \ dx \\\\\implies\displaystyle \sf I = \int \dfrac{cos\ u}{3} du + \int 3sin\ x \ dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3}\int \dfrac{cos(u)}{3} + \int 3sin(x) dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3} sin(u) + C +\int 3sin(x) dx \\\\\implies\displaystyle \sf I = \dfrac{1}{3}sin(u) + C + 3\int sin(x) \ dx \\\\\implies\displaystyle\sf I =  \dfrac{1}{3}sin(u) + C + 3(-cos(x)+C) \\\\\implies \underset{\blue{\sf Required\ Answer }}{\underbrace{\boxed{\boxed{\displaystyle\red{\sf I =  \dfrac{1}{3}sin(3x)  - 3cos(x) + C }}}}}

6 0
3 years ago
Given the Arithmetic series A1+A2+A3+A4 7 + 11 + 15 + 19 + . . . + 91 What is the value of sum?
aivan3 [116]

Answer:

The value of the sum is 1078

Step-by-step explanation:

* Lets revise how to find the sum of the arithmetic series

- In the arithmetic series there is a constant difference between each

 two consecutive terms

- Ex:

# 4 , 9 , 14 , 19 , 24 , .......... (+ 5)

# 25 , 15 , 5 , -5 , -15 , .......... (-10)

- So if the first term is a and the constant difference between each two

  consecutive terms is d, then

  U1 = a , U2 = a + d , U3 = a + 2d , U4 = a + 3d , ..........

- Then the nth term is Un = a + (n - 1)d

- The sum of n terms is Sn = n/2[a + L] , where L is the last term in

 the series

* Lets solve the problem

∵ The arithmetic series is 7 + 11 + 15 + 19 + ......................... + 91

∴ The first term (a) is 7

∴ The last term is (L) 91

- Lets find the constant difference

∵ 11 - 7 = 4 and 15 - 11 = 4

∴ The constant difference (d) is 4

- Lets find the sum of the series from 7 to 91

∵ Sn = n/2[a + L]

∵ a = 7 and L = 91

∴ Sn = n/2[7 + 91]

- We need to know how many terms in the series

∵ L is the last term and equals 91 lets find its position in the series

∵ Un = a + (n - 1)d

∵ a = 7 , d = 4 and Un = 91

∴ 91 = 7 + (n - 1)(4) ⇒ subtract 7 from both sides

∴ 84 = (n - 1)(4) ⇒ divide both sides by 4

∴ 21 = n - 1 ⇒ add 1 to both sides

∴ n = 22

∴ The number of the terms in the series is 22

- Lets find the sum of the 22 terms (S22)

∴ S22 = 22/2[7 + 91]

∴S22 = 11[98] = 1078

* The value of the sum is 1078

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3 years ago
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leva [86]

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Step-by-step explanation:

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6 0
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Vinil7 [7]

Answer:

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6 0
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