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Readme [11.4K]
3 years ago
14

Find the value of this expression if x = 5. x2 – 5

Mathematics
1 answer:
BigorU [14]3 years ago
5 0
The value is 5, the steps above are correct
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Susan solved 200-91 and decided to add her answer to 91 to check her work explain why this strategy works
zhuklara [117]
This strategy works because if you subtract one number from another the sum of the two numbers should be the that is being subtracted from. (In this case, it's 200)
3 0
3 years ago
Read 2 more answers
Which of the following is an equivalent form of the compound inequality
blondinia [14]

Answer:

pls mark brainy

Step-by-step explanation:

option (d) is correct.

An equivalent form of the  given compound inequality −44 > −2x − 8 ≥ −8 is  −44 > −2x − 8 and −2x − 8 ≥ −8

Step-by-step explanation:

Given a compound inequality −44 > −2x − 8 ≥ −8

We have to write  an equivalent form of compound inequality.

Compound inequality  consists of two inequalities joined together and the solution is the intersection of each inequality.

Compound inequality has two sides the left hand side and right hand side we can solve them by taking each inequality one at a time.

For given compound inequality, −44 > −2x − 8 ≥ −8

we have

Left side of inequality as  −44 > −2x − 8

and right side of inequality as  −2x − 8 ≥ −8

Thus, option (d) is correct.

Thus, An equivalent form of the  given compound inequality −44 > −2x − 8 ≥ −8 is  −44 > −2x − 8 and −2x − 8 ≥ −8

3 0
3 years ago
Can somebody help please and thank you :)
Marizza181 [45]
I can help soo first u need to read the question and see what’s its telling u then u got to see what to did like division adding and the other stuff but u are going to divide
4 0
3 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
9- (-2)<br><br><br>Do I add or use KFC? ​
Iteru [2.4K]

Answer:

the answer is 11

you need to add 9 with 2

i think the KFC thecnique cant be use in this question

3 0
3 years ago
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