Please help!!

Sum of 6 divided by a number and that number divided by 6

Korvikt [17]

The sum is :
6 / x + x / 6 =
= ( 6 * 6 ) + ( x * x ) / 6 x =
= ( 36 + x² ) / ( 6 x )
Answer: ( 36 + x² ) ( 6 x )

7 0

3 years ago

Number sequence write the correct number in between to complete the number sequence.

Alborosie

Answer:

1. 15, 13

2. 12, 6

3. 130,135

4. 44, 41

5. 42, 52

Step-by-step explanation:

1. the answer is 15,13 because the sequence goes with minus 2

2. the answer is 12,6 because the sequence goes with divided by 2

3. the answer is 130,135 because the sequence goes with plus 5

4. the answer is 44, 41 because the sequence goes with minus 3

5. the answer is 42,52 because the sequence goes with plus 10

8 0

3 years ago

How do i solve that question?

yawa3891 [41]

a) The solution of this ordinary differential equation is $y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2 } }$ .

b) The integrating factor for the ordinary differential equation is $-\frac{1}{x}$ .

The particular solution of the ordinary differential equation is $y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$ .

<h3>How to solve ordinary differential equations</h3>

a) In this case we need to separate each variable ( $y, t$ ) in each side of the identity:

$6\cdot \frac{dy}{dt} = y^{4}\cdot \sin^{4} t$ (1)

$6\int {\frac{dy}{y^{4}} } = \int {\sin^{4}t} \, dt + C$

Where $C$ is the integration constant.

By table of integrals we find the solution for each integral:

$-\frac{2}{y^{3}} = \frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32} + C$

If we know that $x = 0$ and $y = 1$ , then the integration constant is $C = -2$ .

The solution of this ordinary differential equation is $y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2 } }$ . $\blacksquare$

b) In this case we need to solve a first order ordinary differential equation of the following form:

$\frac{dy}{dx} + p(x) \cdot y = q(x)$ (2)

Where:

$p(x)$ - Integrating factor
$q(x)$ - Particular function

Hence, the ordinary differential equation is equivalent to this form:

$\frac{dy}{dx} -\frac{1}{x}\cdot y = x^{2}+\frac{1}{x}$ (3)

The integrating factor for the ordinary differential equation is $-\frac{1}{x}$ . $\blacksquare$

The solution for (2) is presented below:

$y = e^{-\int {p(x)} \, dx }\cdot \int {e^{\int {p(x)} \, dx }}\cdot q(x) \, dx + C$ (4)

Where $C$ is the integration constant.

If we know that $p(x) = -\frac{1}{x}$ and $q(x) = x^{2} + \frac{1}{x}$ , then the solution of the ordinary differential equation is:

$y = x \int {x^{-1}\cdot \left(x^{2}+\frac{1}{x} \right)} \, dx + C$

$y = x\int {x} \, dx + x\int\, dx + C$

$y = \frac{x^{3}}{2}+x^{2}+C$

If we know that $x = 1$ and $y = -1$ , then the particular solution is:

$y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$

The particular solution of the ordinary differential equation is $y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$ . $\blacksquare$

To learn more on ordinary differential equations, we kindly invite to check this verified question: brainly.com/question/25731911

3 0

3 years ago

Combine like terms. 6x + 4 – 3x + 5

Ivahew [28]

X = 3

Combine like terms: -4 + -5 = -9
-9 + 6x + -3x = 0
Combine like terms: 6x + -3x = 3x
-9 + 3x =0

Solving -9 + 3x = 0

Solving for variable ‘x’

Moving all terms containing my x to the left, all other terms to the right
Add ‘9’ to each side of the equation -9 + 9 + 3x = 0 + 9

Combine like terms: -9 + 9 = 0
0+3x=0+9
3x = 0 + 9

Combine like terms: 0 + 9 = 9
3x = 9

Divide each side by ‘3’.
X = 3

SORRY THIS IS SO LONG

7 0

3 years ago

In one grade of 150 students, the ratio of boys to girls is 2 : 3.

denis23 [38]

Answer:

The numbers of boys in one grade are 60 .

Step-by-step explanation:

As given

In one grade of 150 students .

The ratio of boys to girls is 2:3.

Let us assume that the scalar multiple of boys and girls be x.

Thus number of boys = 2x

Number of girls = 3x

Than the equations becomes

2x + 3x = 150

5x = 150

x = 30

Thus number of boys = 2 × 30

= 60

Therefore the numbers of boys in one grade are 60

8 0

3 years ago

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