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Lerok [7]
3 years ago
12

X in 5xy = 10 to solve the system below. x = 25 - y 5x + y =10

Mathematics
2 answers:
ziro4ka [17]3 years ago
5 0

Answer:

5x+ y = 10   ---(eq1)

x = 25 - y    ---(eq2)

substituting (eq2) in (eq1)

5(25 - y) + y = 10

125 - 5y + y = 10

125 - 4y = 10

-4y = 10 - 125

-4y = -115

4y = 115

y = 115/4 = 28.75

substituting value of y in (eq2)

x = 25 - 28.75

x = -3.75

lora16 [44]3 years ago
4 0

Answer:

Step-by-step explanation:

there are two ways to solve systems of equations,  elimination or substitution, this looks like elimination would work quickly

x= 25 -y

x + y = 25   (eq 1)

5x + y = 10  (eq 2)

subtract 2 from 1

x + y = 25  

-(5x + y = 10)

-4x + 0y = 15

-4x = 15

x = - 15/4

now plug in x into either starting equations and solve for y

-15/4 + y = 25

y = 25 + 15/4

y = 100/4 + 15 /4

y = 115 /4

y = 28 3/4

we have solved it :)

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Arlecino [84]

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C) 4x^2 -8x +4[/tex]

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8 0
3 years ago
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Murljashka [212]

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

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\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 }}}}}

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2 years ago
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