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Harlamova29_29 [7]
3 years ago
10

Help

Mathematics
1 answer:
konstantin123 [22]3 years ago
5 0

Answer:

(2/3)(x-2)=4x

Simplify the left side of the equation.

(2/3)x - (2/3)*2 = 4x

Simplify the second part of the left side.

(2/3)x - (4/3) = 4x

Subtract (2/3)x on each side.

-(4/3) = (10/3)x

Divide by (10/3) on each side.

-0.4 = x

please mark brainliest

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F(x) = x2 − x − ln(x) (a) find the interval on which f is increasing. (enter your answer using interval notation.) find the inte
Mekhanik [1.2K]

Answer:

(a) Decreasing on (0, 1) and increasing on (1, ∞)

(b) Local minimum at (1, 0)

(c) No inflection point; concave up on (0, ∞)

Step-by-step explanation:

ƒ(x) = x² - x – lnx

(a) Intervals in which ƒ(x) is increasing and decreasing.

Step 1. Find the zeros of the first derivative of the function

ƒ'(x) = 2x – 1 - 1/x = 0

           2x² - x  -1 = 0

     ( x - 1) (2x + 1) = 0

         x = 1 or x = -½

We reject the negative root, because the argument of lnx cannot be negative.

There is one zero at (1, 0). This is your critical point.

Step 2. Apply the first derivative test.

Test all intervals to the left and to the right of the critical value to determine if the derivative is positive or negative.

(1) x = ½

ƒ'(½) = 2(½) - 1 - 1/(½) = 1 - 1 - 2 = -1

ƒ'(x) < 0 so the function is decreasing on (0, 1).

(2) x = 2

ƒ'(0) = 2(2) -1 – 1/2 = 4 - 1 – ½  = ⁵/₂

ƒ'(x) > 0 so the function is increasing on (1, ∞).

(b) Local extremum

ƒ(x) is decreasing when x < 1 and increasing when x >1.

Thus, (1, 0) is a local minimum, and ƒ(x) = 0 when x = 1.

(c) Inflection point

(1) Set the second derivative equal to zero

ƒ''(x) = 2 + 2/x² = 0

             x² + 2 = 0

                   x² = -2

There is no inflection point.

(2). Concavity

Apply the second derivative test on either side of the extremum.

\begin{array}{lccc}\text{Test} & x < 1 & x = 1 & x > 1\\\text{Sign of f''} & + & 0 & +\\\text{Concavity} & \text{up} & &\text{up}\\\end{array}

The function is concave up on (0, ∞).

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3 years ago
Whats 4 to the 4th power perenthasis 4
Vinil7 [7]

65,536

you multiple 4 times 4 then you multiply 16 ties 16 times 16 times 16

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PLEASE HELP!!!!<br> will mark brainliest!!
Mariulka [41]

Answer: f(x) = x - 1

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Gente me ajudem prfvv em matemática
Tasya [4]

Answer:

Step-by-step explanation:

Trace a linha de 8cm, daí você vai dividir ela no meio, ou seja, no 4. Aí é só contar 2 cm para cada lado, que vai ser o 1, as pontas vão ser o 2. Tendeu? Daí é só marcar os pontos no -3/7(~-2,3), 1,6, 7/5(1,4), -1 e 0

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2 years ago
What’s the gcf of 18k and 15k cubed?
Contact [7]

Answer:

The GCF for the variable part is  

k

Step-by-step explanation:

Since  

18

k

,

15

k

3

contain both numbers and variables, there are two steps to find the GCF (HCF). Find GCF for the numeric part then find GCF for the variable part.

Steps to find the GCF for  

18

k

,

15

k

3

:

1. Find the GCF for the numerical part  

18

,

15

2. Find the GCF for the variable part  

k

1

,

k

3

3. Multiply the values together

Find the common factors for the numerical part:

18

,

15

The factors for  

18

are  

1

,

2

,

3

,

6

,

9

,

18

.

Tap for more steps...

1

,

2

,

3

,

6

,

9

,

18

The factors for  

15

are  

1

,

3

,

5

,

15

.

Tap for more steps...

1

,

3

,

5

,

15

List all the factors for  

18

,

15

to find the common factors.

18

:  

1

,

2

,

3

,

6

,

9

,

18

15

:  

1

,

3

,

5

,

15

The common factors for  

18

,

15

are  

1

,

3

.

1

,

3

The GCF for the numerical part is  

3

.

GCF

Numerical

=

3

Next, find the common factors for the variable part:

k

,

k

3

The factor for  

k

1

is  

k

itself.

k

The factors for  

k

3

are  

k

⋅

k

⋅

k

.

k

⋅

k

⋅

k

List all the factors for  

k

1

,

k

3

to find the common factors.

k

1

=

k

k

3

=

k

⋅

k

⋅

k

The common factor for the variables  

k

1

,

k

3

is  

k

.

k

The GCF for the variable part is  

k

.

GCF

Variable

=

k

Multiply the GCF of the numerical part  

3

and the GCF of the variable part  

k

.

3

k

3 0
3 years ago
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