All categories

2 years ago

Plz help, photo above

Mathematics

2 answers:

Kitty [74]2 years ago

3 0

Answer:

Step-by-step explanation:

Can I have brainliest? It would help me out, if not thanks anyways! Hope this helped and have a nice day!

blsea [12.9K]2 years ago

3 0

D hope this helps you

You might be interested in

Find the equation of the line. Use exact numbers <br><br> Y= __ x+ __

nika2105 [10]

Answer:

y = 3/4x -2

Step-by-step explanation:

slope = rise/run = 3/4

y intercept is where the line crosses the y axis, it crosses at -2

5 0

2 years ago

An investor invested a total of $2500 in two mutual funds. One fund earned a 5% profit while the other earned a 3% profit. If th

7nadin3 [17]

Answer:

$500 and $2000

Step-by-step explanation:

Let x represent the total investment = $2500

also, this total is split into two different funds

Lets represent these funds as a and b, such that fund a yields a profit of 3% and fund b yield a profit of 5%

So,

a + b = x

a + b = 2500 ......eq 1

Profit from each fund gives;

0.03 a + 0.05b = 115 ....eq 2

Solve simultaneously using substitution method

From eq 1;

b = 2500 - a

Slot in this value in eq 2

0.03a + 0.05 (2500 - a) = 115

expand

0.03a + 125 - 0.05a = 115

collect like terms

0.03a - 0.05a = 115 - 125

-0.02a = -10

Divide both sides by -0.02

a = $500

Put this value of a in eq 1

500 + b = 2500

Subtract 500 from both sides

b = 2500 - 500

b = $2000

3 0

2 years ago

Find a particular solution to y" - y + y = 2 sin(3x)

leonid [27]

Answer with explanation:

The given differential equation is

y" -y'+y=2 sin 3x------(1)

Let, y'=z

y"=z'

$\frac{dy}{dx}=z\\\\d y=zdx\\\\y=z x$

Substituting the value of , y, y' and y" in equation (1)

z'-z+zx=2 sin 3 x

z'+z(x-1)=2 sin 3 x-----------(1)

This is a type of linear differential equation.

Integrating factor

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

Multiplying both sides of equation (1) by integrating factor and integrating we get

$\rightarrow z\times e^{\frac{x^2}{2}-x}=\int 2 sin 3 x \times e^{\frac{x^2}{2}-x} dx=I$

$I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx -\int \frac{2\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx-\frac{2I}{3}\\\\\frac{5I}{3}=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{5}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{5} dx$