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3 years ago

Solve the following equation

Mathematics

1 answer:

Ilya [14]3 years ago

7 0

Answer:

x = 5

Step-by-step explanation:

here's the solution in attached image

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Evaluate the expression. Will choose brainliest

svet-max [94.6K]

Answer:

-213

Step-by-step explanation:

3|3-5|^2-(-15)^2

3(4)-225

=12-225= -213

3 0

3 years ago

Read 2 more answers

Solve the initial value problems.

slavikrds [6]

Both equations are linear, so I'll use the integrating factor method.

The first ODE

$xy' + (x+1)y = 0 \implies y' + \dfrac{x+1}x y = 0$

has integrating factor

$\exp\left(\displaystyle \int\frac{x+1}x \, dx\right) =\exp\left(x+\ln(x)\right) = xe^x$

In the original equation, multiply both sides by eˣ :

$xe^x y' + (x+1) e^x y = 0$

Observe that

d/dx [xeˣ] = eˣ + xeˣ = (x + 1) eˣ

so that the left side is the derivative of a product, namely

$\left(xe^xy\right)' = 0$

Integrate both sides with respect to x :

$\displaystyle \int \left(xe^xy\right)' \, dx = \int 0 \, dx$

$xe^xy = C$

Solve for y :

$y = \dfrac{C}{xe^x}$

Use the given initial condition to solve for C. When x = 1, y = 2, so

$2 = \dfrac{C}{1\cdot e^1} \implies C = 2e$

Then the particular solution is

$\boxed{y = \dfrac{2e}{xe^x} = \dfrac{2e^{1-x}}x}$

The second ODE

$(1+x^2)y' - 2xy = 0 \implies y' - \dfrac{2x}{1+x^2} y = 0$

has integrating factor

$\exp\left(\displaystyle \int -\frac{2x}{1+x^2} \, dx\right) = \exp\left(-\ln(1+x^2)\right) = \dfrac1{1+x^2}$

Multiply both sides of the equation by 1/(1 + x²) :

$\dfrac1{1+x^2} y' - \dfrac{2x}{(1+x^2)^2} y = 0$

and observe that

d/dx[1/(1 + x²)] = -2x/(1 + x²)²

Then

$\left(\dfrac1{1+x^2}y\right)' = 0$

$\dfrac1{1+x^2}y = C$

$y = C(1 + x^2)$

When x = 0, y = 3, so

$3 = C(1+0^2) \implies C=3$

$\implies \boxed{y = 3(1 + x^2) = 3 + 3x^2}$

7 0

2 years ago

Solve for z -0.25z= -1.25

GarryVolchara [31]

Answer:

Step-by-step explanation:

-0.25z= -1.25

Convert the decimals to improper fractions

That's

So we have

Multiply through by 4

We have

- z = - 5

Divide both sides by - 1

the final answer is

Hope this helps you

6 0

3 years ago

Read 2 more answers

Which fraction is equivalent to <img src="https://tex.z-dn.net/?f=%5Cfrac%7B21%7D%7B28%7D" id="TexFormula1" title="\frac{21}{28}

BaLLatris [955]

$\dfrac{21}{28}=\dfrac{21:7}{28:7}=\dfrac{3}{4}\to\boxed{C}$

5 0

3 years ago

A rectangular field is 16m long and 10m wide

Llana [10]

Answer:

path =2m

Step-by-step explanation:

6 0

4 years ago

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Solve the following equation​

Solve the following equation