Is y=6, y=3x+4 consistent or inconsistent

Talja [164]

I did not get any zeros since the graph doesn’t cross the x axis, meaning that there are no rational zeros

However, here is the method u can use to find the zeros lol

You can use the quadratic formula in order to get the zeros

This is the equation therefore use these values
ax^2+bx+c=0

A=1
B= -5
C=12

The quadratic formula is -b±√(b^2-4ac))/2a (I left a picture just in case)

6 0

3 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

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

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

8 0

3 years ago

Qtr is a right triangle TQR is a right triangle QS is an altitude which statements are true

Artemon [7]

Answer:

the answer is 1and 3 that's right

4 0

3 years ago

Helppp will give brainless 35points

Ugo [173]

Answer:

$\boxed{question \: number \: \to 6} \\ \boxed{ w = 72} \\ \boxed{ x =144 } \\ \boxed{ y = 108}$

Step-by-step explanation:

$y + w + y + w = 360 \to \sum \: of \: angles \\ y = \: \boxed{opp \: y} \to \: vertically \: opposite \: angles \: are \: equal \\ w = \: \boxed{opp \: w} \to \: the \: same \: reason \: as \: y\\ hence \to \\ 2(w+ y) = 360 \\ w + y = 180 \\ but \: 5w = 360 \to \: angles \: at \: a \: point \\ w = \frac{360}{5} \\ \boxed{w = 72} \\ therefore \to \\ w + y = 180 \\ 72 + y = 180 \\ y = 180 - 72 \\ \boxed{ y = 108} \\ hence \: \boxed{x} \: is \: given \: by \to \\y + x + y = 360 \to \: angles \: at \: a \: point \\108 + x + 108 = 360 \\ x = 360 - 216 \\ \boxed{ x = 144}$

♨Rage♨............You can also ask me the other questions separatly ........cos i cant cover all in this one.♨

6 0

3 years ago

Read 2 more answers

Evaluate the expression if H=120 and T=12.

dangina [55]

what is the expression

7 0

4 years ago