One degree Celsius indicates the same temperature change as

Two non-common sides of adjacent supplementary angles form a _____ angle.

mel-nik [20]

Answer:

C. straight

Step-by-step explanation:

A Linear Pair is two adjacent angles whose non-common sides form opposite rays.

If two angles form a linear pair, the angles are supplementary.

A linear pair forms a straight angle which contains 180º, so you have 2 angles whose measures add to 180, which means they are supplementary.

In the figure given in attachment, AB and BC are two non common sides of ∠ABD and ∠DBC.

∠1 and ∠2 form a linear pair.

The line through points A, B and C is a straight line.

∠1 and ∠2 are supplementary.

Thus two non-common sides of adjacent supplementary angles form a <u>straight</u> angle.

8 0

2 years ago

Apply the method of undetermined coefficients to find a particular solution to the following system.wing system.

jarptica [38.1K]

$y''-y'+y=\sin x$

The corresponding homogeneous ODE has characteristic equation $r^2-r+1=0$ with roots at $r=\dfrac{1\pm\sqrt3}2$ , thus admitting the characteristic solution

$y_c=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x$

For the particular solution, assume one of the form

$y_p=a\sin x+b\cos x$

${y_p}'=a\cos x-b\sin x$

${y_p}''=-a\sin x-b\cos x$

Substituting into the ODE gives

$(-a\sin x-b\cos x)-(a\cos x-b\sin x)+(a\sin x+b\cos x)=\sin x$

$-b\cos x+a\sin x=\sin x$

$\implies a=1,b=0$

Then the general solution to this ODE is

$\boxed{y(x)=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x+\sin x}$

$y''-3y'+2y=e^x\sin x$

$\implies r^2-3r+2=(r-1)(r-2)=0\implies r=1,r=2$

$\implies y_c=C_1e^x+C_2e^{2x}$

Assume a solution of the form

$y_p=e^x(a\sin x+b\cos x)$

${y_p}'=e^x((a+b)\cos x+(a-b)\sin x)$

${y_p}''=2e^x(a\cos x-b\sin x)$

Substituting into the ODE gives

$2e^x(a\cos x-b\sin x)-3e^x((a+b)\cos x+(a-b)\sin x)+2e^x(a\sin x+b\cos x)=e^x\sin x$

$-e^x((a+b)\cos x+(a-b)\sin x)=e^x\sin x$

$\implies\begin{cases}-a-b=0\\-a+b=1\end{cases}\implies a=-\dfrac12,b=\dfrac12$

so the solution is

$\boxed{y(x)=C_1e^x+C_2e^{2x}-\dfrac{e^x}2(\sin x-\cos x)}$

$y''+y=x\cos(2x)$

$r^2+1=0\implies r=\pm i$

$\implies y_c=C_1\cos x+C_2\sin x$

Assume a solution of the form

$y_p=(ax+b)\cos(2x)+(cx+d)\sin(2x)$

${y_p}''=-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x)$

Substituting into the ODE gives

$(-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x))+((ax+b)\cos(2x)+(cx+d)\sin(2x))=x\cos(2x)$

$-(3ax+3b-4c)\cos(2x)-(3cx+3d+4a)\sin(2x)=x\cos(2x)$

$\implies\begin{cases}-3a=1\\-3b+4c=0\\-3c=0\\-4a-3d=0\end{cases}\implies a=-\dfrac13,b=c=0,d=\dfrac49$

so the solution is

$\boxed{y(x)=C_1\cos x+C_2\sin x-\dfrac13x\cos(2x)+\dfrac49\sin(2x)}$

7 0

3 years ago

Please help I don't understand what to do

harina [27]

Answer:

This is an a fake answer but I want to supposed to be multiple pictures?? Because you said add an explanation for each picture and is this history? Are you sure it’s math

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Ana escribe conjuntos de cinco enteros positivos consecutivos con la siguiente propiedad: La suma de tres de los números es tan

Keith_Richards [23]

Answer:

a) 1.

Step-by-step explanation:

Supongamos que el primer número entero positivo es a tal que a ∈ Z⁺.

Luego, el conjunto de los cinco enteros positivos se puede expresar como

A = {a; a+1; a+2; a+3; a+4}

Dada la condición del problema, se debe cumplir que

a + (a+1) + (a+2) = (a+3) + (a+4)

⇒ 3a + 3 = 2a + 7

Resolviendo la ecuación resulta

a = 4

Luego, el conjunto A nos resulta

A = {4; 5; 6; 7; 8}

Puede concluirse que sólo un conjunto cumple con esta condición.

5 0

2 years ago

How do change 0.9 in fraction

Fantom [35]

0.9 as a fraction is: $\frac{9}{10}$

What place is the '9' in? It's in the tenth place.
0.9 can be said as $nine \ tenths$

Nine $(9)$ tenths $(10)$ = $\frac{9}{10}$

3 0

3 years ago

Read 2 more answers