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3 years ago

Which value is a solution of the equation 3-2x=-5 A= 2 B= 4 C= -2 D= 1/5

Mathematics

2 answers:

Ad libitum [116K]3 years ago

6 0

3 - 2x = -5
-2x = -8
x = 4

krok68 [10]3 years ago

5 0

B=4 -5 - 3 = -8 -8/ -2 = 4

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Solve the linear differential equation 2xy' + y = 2√x

77julia77 [94]

Answer:

Step-by-step explanation:

General form of the linear differential equation can be written as:

$\frac{dy}{dx}+P(x)y=Q(x)$

For this case, we can rewrite the equation as:

$\frac{dy}{dx}+\frac{1}{2x}y=\frac{\sqrt{x}}{x}$

Here $P(x) =\frac{1}{2x}; Q(x)=\frac{\sqrt{x}}{x}$

To find the solution (y(x)), we can use the integration factor method:

$Fy(x)=\int Q(x)Fdx+C \rightarrow F=e^{\int P(x)dx$

Then $F=e^{\int \frac{1}{2x}dx}=e^{\frac{1}{2}\ln|x|\right}=\sqrt{|x|}$

So, we can find:

$y\sqrt{|x|}=\int \frac{\sqrt{x}\sqrt{|x|}}{x}dx+C$

Suppose that $x\in \double R$ , then $\sqrt{|x|}=\sqrt{x}$ , and we find:

$y\sqrt{x}=x+C \rightarrow y(x)=\sqrt{x}+\frac{C\sqrt{x}}{x}$

To check our solution is right or not, put your y(x) back to the ODE:

$y' = \frac{1}{2\sqrt{x}}-\frac{C}{2\sqrt{x^{3}}}$

$2xy'=\frac{x-C}{\sqrt{x}}$

$2xy'+y=\frac{x-C}{\sqrt{x}}+\sqrt{x}+\frac{C\sqrt{x}}{x}=2\sqrt{x}$

(it means your solution is right)

3 0

2 years ago

What is x+2y=10 when x=2

Gwar [14]

Answer: y=4

Step-by-step explanation:

x=2 so the equation would read 2+2y=10 and u want to get y alone so first u subtract by 2 on both sides so 2-2+2y=10-2 which is equal to 2y=8 so now u divide by 2 on both sides so it is y=4

7 0

4 years ago

Read 2 more answers

There are 18 girls in a class and the ratio of the number of girls and boys in the class is 3:2. Find the number of boys.

Paraphin [41]

Answer:

Step-by-step explanation:

Multiply to get the number of girls they've given you.

3 x 6 = 18 girls.

2 x 6 = 12 boys.

5 0

3 years ago

Read 2 more answers

1. -10=xy+z, solve for x please? 2. h-4/j =k, solve for j please?

ANEK [815]

Answer:

$x=\frac{-10-z}{y}$

Step-by-step explanation:

1 Subtract z from both sides.

$-10-z=xy$

2 Divide both sides by y

$\frac{-10-z}{y} =x$

3 Switch sides.

$x=\frac{-10-z}{y}$

2nd question

Answer: $j=-\frac{4}{k-h}$

Step-by-step explanation:

1 Subtract h from both sides

$-\frac{4}{j} =k-h$

2 Multiply both sides by j.

$-4=(k-h)j$

3 Divide both sides by k−h.

$-\frac{4}{k-h}=j$

4 Switch sides.

$j=-\frac{4}{k-h}$

8 0

4 years ago

Read 2 more answers

Last year, sales at Company X were 10% greater in February than in January, 15% less in March than in Feb, 20% greater in April

horsena [70]

Answer:

May

Step-by-step explanation:

Let x be the sales in January,

∵ Sales were 10% greater in February than in January,

So, the sales in February = (100 +10)% of x

= 110% of x

= $\frac{110x}{100}$

= 1.1x

∵ it is 15% less in March than in Feb,

Sales in march = (100-15)% sales in February

= 85% of 1.1x

= 0.935x

∵ It is 20% greater in April than in March,

Sales in April = (100+20)% of sales in march

= 120% of 0.935x

= 1.122x,

∵ It is 10% less in May than in April

Sales in May = (100-10)% of sales in march

= 90% of 1.122x,

= 1.0098x,

∵ It is 5% greater in June than in May

Sales in June = (100+5)% of sales in march

= 105% of 1.0098x,

= 1.5147x

∵ 1.0098x is much closure to x than 1.1x, 0.935x, 1.122x and 1.5147x

Hence, the sales in May is closest to sales in January.

6 0

4 years ago