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Alenkinab [10]
3 years ago
5

PLEASE HELP! 15 POINTS TO BRAINLIEST!!!

Mathematics
1 answer:
Semenov [28]3 years ago
5 0

first one no second one no third one no frouth no fifth yes last one no

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Create a table and graph the equation y =2x-2
soldi70 [24.7K]

Table:

x  0,1

y  -2,0

Graph is the attached image.

7 0
2 years ago
In the inequality "y is less than or equal to 5 and is greater than or equal to -2," the product of the endpoints will be which
Licemer1 [7]

Answer:

  • (C) negative

Step-by-step explanation:

<u>Inequality "y is less than or equal to 5 and is greater than or equal to -2":</u>

  • - 2 ≤ y ≤ 5

<u>This is the interval:</u>

  • y ∈ [- 2, 5]

<u>The endpoints are:</u>

  • - 2 and 5

<u>Product of the endpoints:</u>

  • - 2 × 5 = - 10

<u>Answer choices:</u>

  • (A) odd ⇒ False, - 10 is even
  • (B) positive ⇒ False, - 10 is not positive
  • (C) negative ⇒ True
  • (D) None of the above​ ⇒ False, option C is correct
3 0
2 years ago
Read 2 more answers
About 20​% of babies born with a certain ailment recover fully. a hospital is caring for sixsix babies born with this ailment. t
alisha [4.7K]

Answer :

Yes, the experiment is a binomial experiment because it satisfies all the four characteristics of the binomial experiment :

1. Fixed number of trials

2. Exactly two outcomes

3. Outcomes are independent of each other

4. Probability of all the outcomes remains constant from trial to trial.

Success in the given experiment is that the baby recovers fully.

Hospital is caring for 6 babies , n = 6

p = 0.20 ( Success )

q = 1 - 0.20 = 0.80 ( Not a success )

Value of random variable x is the number of babies which are recovering fully, x = 0, 1, 2, 3, 4, 5 or 6

5 0
3 years ago
Look at the picture
valentina_108 [34]

9514 1404 393

Answer:

  A

Step-by-step explanation:

When angles are between the parallel lines, they are called <em>interior</em>. If they are both on the same side of the transversal, they are "same-side interior."

If angles are on opposite sides of the transversal, they are <em>exterior</em>.

If angles are in the same direction from the intersection point, they are <em>corresponding</em>.

Angles in opposite directions from the point of intersection are <em>vertical</em> angles.

  same-side interior: {3, 5} and {4, 6}. (supplementary)

  same-side exterior: {1, 7} and {2, 8}. (supplementary)

  alternate interior: {3, 6} and {4, 5}. (congruent)

  alternate exterior: {1, 8} and {2, 7}. (congruent)

  corresponding: {1, 5}, {2, 6}, {3, 7}, {4, 8}. (congruent)

  vertical: {1, 4}, {2, 3}, {5, 8}, {6, 7}. (congruent)

Angles 6 and 4 are same-side interior angles; their sum is 180°.

6 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
2 years ago
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