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3 years ago

What is an estimate of the solution. If the equation 6n + 3 = 2? Use a table.

1 answer:

4 0

I don't know how to use the table but to solve the problem:

1. Subtract 3 from both sides
6n=2-3

2. Simplify 2-3 to -1
6n=-1

3. Divide both sides by 6
n=1/6

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Solving the equation of <br> 7x-5+5x=15x-7-3x+2

Masja [62]

Answer: no solution

Step-by-step explanation:

7x + 5 + 5x = 15x - 7 - 3x + 2

12x + 5 = 12x - 5

-12x -5. -12x -5

0 = -10

no solution

7 0

2 years ago

During the summer Austin sells tomatoes at his family's produce stand . Every morning he starts with 150 tomatoes on. Sunday Aus

Marianna [84]

Answer:

30%

Step-by-step explanation:

Given:

Every morning he starts with 150 tomatoes.

On Sunday Austin sells 45 of the 150 tomatoes.

Question asked:

What percent of the tomatoes he sold ?

Solution:

As Austin sells 45 tomatoes out of 150 tomatoes, we will find percent of the tomatoes he sold by using :

Percent of the tomatoes he sold = Number of tomatoes he sold divided by total number of tomatoes he starts with.

$Percentage =\frac{45}{150} \times100\\$

$=\frac{4500}{150} \\ = 30$

Therefore, 30% of the tomatoes he sold on Sunday.

5 0

3 years ago

Please help mee (don't send a link)

zlopas [31]

Answer:

Step-by-step explanation:

the sign flips when you divide or multiply one

5 0

2 years ago

Read 2 more answers

The probability that a person in a certain town has brown eyes is 2 out of 5. A survey of 450 people from that same town was tak

Annette [7]

Answer:

It could be c

Step-by-step explanation:

180

7 0

3 years ago

Read 2 more answers

Please, I need help in this ??

nignag [31]

Answer:

$\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c$

Step-by-step explanation:

$\int\frac{x^{4}}{x^{4} -1}dx$

Adding and Subtracting 1 to the Numerator

$\int\frac{x^{4} - 1 + 1}{x^{4} -1}dx$

Dividing Numerator seperately by $x^{4} - 1$

$\int 1 + \frac{1}{x^{4}-1 }\, dx$

Here integral of 1 is x +c1 (where c1 is constant of integration

$x + c1 + \int\frac{1}{(x-1)(x+1)(x^{2}+1)}\, dx$ ----------------------------------(1)

We apply method of partial fractions to perform the integral

$\frac{1}{(x-1)(x+1)(x^{2}+1)}$ = $\frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{x^{2} + 1}$ ------------------------------------------(2)

$\frac{1}{(x-1)(x+1)(x^{2}+1)} = \frac{A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)}{(x-1)(x+1)(x^{2} +1)}$

1 = $A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)$ -------------------------(3)

Substitute x= 1 , -1 , i in equation (3)

1 = A(1+1)(1+1)

A = $\frac{1}{4}$

1 = B(-1-1)(1+1)

B = $-\frac{1}{4}$

1 = C(i-1)(i+1)

C = $-\frac{1}{2}$

Substituting A, B, C in equation (2)

$\int\frac{x^{4}}{x^{4} -1}dx$ = $\int\frac{1}{4(x-1)} - \frac{1}{4(x+1)} -\frac{1}{2(x^{2}+1) }$

On integration

Here $\int \frac{1}{x}dx = lnx and \int\frac{1}{x^{2}+1 } dx = arctanx$

$\int\frac{x^{4}}{x^{4} -1}dx$ = $\frac{1}{4} ln(x-1)$ - $\frac{1}{4} ln(x+1)$ - $\frac{1}{2} arctanx$ + c2---------------------------------------(4)

Substitute equation (4) back in equation (1) we get

$x + c1 + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1) - \frac{1}{2} arctanx + c2$

Here c1 + c2 can be added to another and written as c

Therefore,

$\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c$

4 0

3 years ago