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olga_2 [115]
3 years ago
10

( Solving Linear Equations in One Variable )

Mathematics
2 answers:
astraxan [27]3 years ago
4 0

Answer:

x=-2

Step-by-step explanation:

Simplify parentheses : 0.5x+0.3=2.1x+9-5.5

Combine like terms : 0.5x+0.3=2.1x+3.5

Isolate the variable : -1.6x=3.2

Divide by -1.6 on both sides : x=-2

allsm [11]3 years ago
3 0

Answer:

x=-2

Step-by-step explanation:

1/4(2x + 1.2) = 3(0.7x + 3) - 5 1/2

multiply both sides of the equation by 4 to get rid of the fraction 1/4

2x+1.2=12(0.7x+3)-4*5 1/2

distribute 12 into parenthesis

2x+1.2= 8.4x+36-4*5 1/2

multiply -4 with 5 1/2

2x+1.2=8.4x+36-22

subtract numbers

2x+1.2= 8.4x+14

move X's to one side and numbers to another

2x-8.4x=14-1.2

collect like terms

-6.4x=12.8

divide by -6.4 to get x

x= -2

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The inequality describing the interval is thusly B,

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Step-by-step explanation:

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I need help with part "C" and "D"
dmitriy555 [2]

3x²cos( x³ ) and 3sin²( x ) cos( x ) are the derivatives of the composite functions f(x) = sin(x³) and f(x) = sin³(x) respectively.

<h3>What are the derivative of f(x) = sin(x³) and f(x) = sin³(x)?</h3>

Chain rule simply shows how to find the derivative of a composite function. It states that;

d/dx[f(g(x))] = f'(g(x))g'(x)

Given the data in the question;

  • f(x) = sin(x³) = ?
  • f(x) = sin³(x) = ?

First, we find the derivate of the composite function f(x) = sin(x³) using chain rule.

d/dx[f(g(x))] = f'(g(x))g'(x)

f(x) = sin(x)

g(x) = x³

Apply chain rule, set u as x³

d/du[ sin( u )] d/dx[ x³ ]

cos( u ) d/dx[ x³ ]

cos( x³ ) d/dx[ x³ ]

Now, differentiate using power rule.

d/dx[ xⁿ ] is nxⁿ⁻¹

cos( x³ ) d/dx[ x³ ]

In our case, n = 3

cos( x³ ) ( 3x² )

Reorder the factors

3x²cos( x³ )

Next, we find the derivative of f(x) = sin³(x)

d/dx[f(g(x))] = f'(g(x))g'(x)

f( x ) = x³

g( x ) = sin( x )

Apply chain rule, set u as sin( x )

d/du[ u³ ] d/dx[ sin( x )]

Now, differentiate using power rule.

d/dx[ xⁿ ] is nxⁿ⁻¹

d/du[ u³ ] d/dx[ sin( x )]

3u²  d/dx[ sin( x )]

Replace the u with sin( x )

3sin²(x)  d/dx[ sin( x )]
Derivative of sin x with respect to x is cos (x)

3sin²( x ) cos( x )

Therefore, the derivatives of the functions are 3x²cos( x³ ) and 3sin²( x ) cos( x ).

Learn more about chain rule here: brainly.com/question/2285262

#SPJ1

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saw5 [17]

Answer:

b, d, and e

Step-by-step explanation:

not a because anything to 0th power is 0 of undefined

b because when you multiply you add the exponents together

d because when you have an exponent on the outside you distribute it and multiply

d because when you divide you subtract and -- is the same as /

3 0
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