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3 years ago

What the three missing terms after22,28,21,27,20,26,19

Mathematics

1 answer:

Tanzania [10]3 years ago

7 0

Answer:

25, 18, 24

Step-by-step explanation:

the pattern is up 6, down 7, so 19+6=25, then 25-7=18 and finally 18+6=24

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If x plus one upon X equal to 11 find the value of x square + 1 upon x square

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119 <answer (check the solution in above pic)

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3 years ago

Surveys conducted in American high schools concluded that 90% of the students in a sample of 400 students had more than one acti

kenny6666 [7]

The sample proportion has an approximately normal distribution, with the mean being the population mean and the standard deviation (i.e. the standard error of the sampling proportion) equal to

SE = sqrt(p*q/n) where p is the population proportion q is 1-p, and n is the sample size.

<span>(Typically, for a confkdence interval, the sample proportion and its complement are substituted for p and q.) </span>

The margin of error will be Zc*SE, where Zc is the critical value of Z for the level of confidence required. The confidence interval, then, will be the sample proportion, plus or minus the margin of error.

For a 90% confidence interval, Zc is 1.645.

<span>So the margin of error is 1.645 * sqrt(.9*.1/400), roughly .025. So your confidence interval would be around [.875, .925]

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2 years ago

Read 2 more answers

What is the sign of -201 + 189−201+189minus, 201, plus, 189?

svlad2 [7]

-201+189=-12-201=-213+189=-24 . S the sign would be negative(-).

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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

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

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2 years ago

Two triangles can be formed with the given information. Use the Law of Sines to solve the triangles.

EastWind [94]

Answer:

The Law of Sines applies to any triangle and works as follows:

a/sinA = b/sinB = c/sinC

We are attempting to solve for every angle and every side of the triangle. With the given information, A = 61°, a = 17, b = 19, we can solve for the unknown angle that is B.

a/sinA = b/sinB

17/sin61 = 19/sinB

sinB = (19/17)(sin61)

sinB = 0.9774

sin-1(sinB) = sin-1(0.9774)

B = 77.8°

With angle B we can solve for angle C and then side c.

A + B + C = 180°

C = 180° - A - B

C = 180° - 61° - 77.8°

C = 41.2°

a/sinA = c/sinC

17/sin61 = c/sin41.2

c = 17(sin41.2/sin61)

c = 12.8

The first solved triangle is:

A = 61°, a = 17, B = 77.8°, b = 19, C = 41.2°, c = 12.8

However, when we solved for angle B initially, that was not the only possible answer because of the fact that sinB = sin(180-B).

The other angle is simply 180°-77.8° = 102.2°. Therefore, angle B can also be 102.2° which will give us different values for c and C.

C = 180° - A - B

C = 180° - 61° - 102.2°

C = 16.8°

a/sinA = c/sinC

17/sin61 = c/sin16.8

c = 17(sin16.8/sin61)

c = 5.6

The complete second triangle has the following dimensions:

A = 61°, a = 17, B = 102.2°, b = 19, C = 16.8°, c = 5.6

The answer you are looking for is the first option given in the question:

B = 77.8°, C = 41.2°, c = 12.8; B = 102.2°, C = 16.8°, c = 5.6

Step-by-step explanation:

8 0

2 years ago

What the three missing terms after22,28,21,27,20,26,19​

What the three missing terms after22,28,21,27,20,26,19