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marissa [1.9K]
3 years ago
7

How to find the first n terms of a quadratic sequence a1 = 3, r = 2, n =5

Mathematics
1 answer:
Nezavi [6.7K]3 years ago
3 0

We suspect you mean "geometric sequence". Multiply each term by r to get the next.

a1 = 3 . . . . . given

a2 = 3*2 = 6

a3 = 6*2 = 12

a4 = 12*2 = 24

a5 = 24*2 = 48

_____

A <em>quadratic</em> sequence is one in which second-differences are a constant. You have not provided enough information about the sequence so that more terms of a quadratic sequence could be found. While you may occasionally see problems asking you to write the formula for one of these, they are rarely studied in any detail in Algebra. At least three terms are needed before the next terms can be predicted.

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Tori is cutting fabric squares to make a quilt. Her squares on average are 5 in. on each side with a standard deviation of 0.1 i
lapo4ka [179]

Answer:

\approx 68\%

Step-by-step explanation:

For normal distributions only, all data falls within approximately 68% of one standard deviation, 95% of two standard deviations, and close to 100% of three standard deviations. The standard deviation is far too small to represent two or three standard deviations, hence \implies \boxed{68\%}.

*Important: This problem would be unsolvable if the question did not say her cuts were normally distributed, because the information above is only applicable to normal distributions.

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3 years ago
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In this problem you will use undetermined coefficients to solve the nonhomogeneous equation y′′+4y′+3y=8te^−t+6e^−t−(9t+6)
Luden [163]

We're given the ODE,

<em>y''</em> + 4<em>y'</em> + 3<em>y</em> = 8<em>t</em> exp(-<em>t </em>) + 6 exp(-<em>t</em> ) - (9<em>t</em> + 6)

(where I denote exp(<em>x</em>) = <em>eˣ </em>)

First determine the characteristic solution:

<em>y''</em> + 4<em>y'</em> + 3<em>y</em> = 0

has characteristic equation

<em>r</em> ² + 4<em>r</em> + 3 = (<em>r</em> + 1) (<em>r</em> + 3) = 0

with roots at <em>r</em> = -1 and <em>r</em> = -3, so the characteristic solution is

<em>y</em> = <em>C</em>₁ exp(-<em>t</em> ) + <em>C</em>₂ exp(-3<em>t</em> )

For the non-homogeneous equation, assume two ansatz solutions

<em>y</em>₁ = (<em>at</em> ² + <em>bt</em> + <em>c</em>) exp(-<em>t </em>)

and

<em>y</em>₂ = <em>at</em> + <em>b</em>

<em />

• <em>y''</em> + 4<em>y'</em> + 3<em>y</em> = 8<em>t</em> exp(-<em>t </em>) + 6 exp(-<em>t</em> ) … … … [1]

Compute the derivatives of <em>y</em>₁ :

<em>y</em>₁ = (<em>at</em> ² + <em>bt</em> + <em>c</em>) exp(-<em>t </em>)

<em>y</em>₁' = (2<em>at</em> + <em>b</em>) exp(-<em>t </em>) - (<em>at</em> ² + <em>bt</em> + <em>c</em>) exp(-<em>t </em>)

… = (-<em>at</em> ² + (2<em>a</em> - <em>b</em>) <em>t</em> + <em>b</em> - <em>c</em>) exp(-<em>t </em>)

<em>y</em>₁'' = (-2<em>at</em> + 2<em>a</em> - <em>b</em>) exp(-<em>t </em>) - (-<em>at</em> ² + (2<em>a</em> - <em>b</em>) <em>t</em> + <em>b</em> - <em>c</em>) exp(-<em>t </em>)

… = (<em>at</em> ² + (<em>b</em> - 4<em>a</em>) <em>t</em> + 2<em>a</em> - 2<em>b</em> + <em>c</em>) exp(-<em>t</em> )

Substitute them into the ODE [1] to get

→   [(<em>at</em> ² + (<em>b</em> - 4<em>a</em>) <em>t</em> + 2<em>a</em> - 2<em>b</em> + <em>c</em>) + 4 (-<em>at</em> ² + (2<em>a</em> - <em>b</em>) <em>t</em> + <em>b</em> - <em>c</em>) + 3 (<em>at</em> ² + <em>bt</em> + <em>c</em>)] exp(-<em>t</em> ) = 8<em>t</em> exp(-<em>t </em>) + 6 exp(-<em>t</em> )

(<em>at</em> ² + (<em>b</em> - 4<em>a</em>) <em>t</em> + 2<em>a</em> - 2<em>b</em> + <em>c</em>) + 4 (-<em>at</em> ² + (2<em>a</em> - <em>b</em>) <em>t</em> + <em>b</em> - <em>c</em>) + 3 (<em>at</em> ² + <em>bt</em> + <em>c</em>) = 8<em>t</em> + 6

4<em>at</em> + 2<em>a</em> + 2<em>b</em> = 8<em>t</em> + 6

→   4<em>a</em> = 8   and   2<em>a</em> + 2<em>b</em> = 6

→   <em>a</em> = 2   and   <em>b</em> = 1

→   <em>y</em>₁ = (2<em>t</em> ² + <em>t </em>) exp(-<em>t </em>)

(Note that we don't find out anything about <em>c</em>, but that's okay since it would have gotten absorbed into the first characteristic solution exp(-<em>t</em> ) anyway.)

• <em>y''</em> + 4<em>y'</em> + 3<em>y</em> = -(9<em>t</em> + 6) … … … [2]

Compute the derivatives of <em>y</em>₂ :

<em>y</em>₂ = <em>at</em> + <em>b</em>

<em>y</em>₂' = <em>a</em>

<em>y</em>₂'' = 0

Substitute these into [2] :

4<em>a</em> + 3 (<em>at</em> + <em>b</em>) = -9<em>t</em> - 6

3<em>at</em> + 4<em>a</em> + 3<em>b</em> = -9<em>t</em> - 6

→   3<em>a</em> = -9   and   4<em>a</em> + 3<em>b</em> = -6

→   <em>a</em> = -3   and   <em>b</em> = 2

→   <em>y</em>₂ = -3<em>t</em> + 2

Then the general solution to the original ODE is

<em>y(t)</em> = <em>C</em>₁ exp(-<em>t</em> ) + <em>C</em>₂ exp(-3<em>t</em> ) + (2<em>t</em> ² + <em>t </em>) exp(-<em>t </em>) - 3<em>t</em> + 2

Use the initial conditions <em>y</em> (0) = 2 and <em>y'</em> (0) = 2 to solve for <em>C</em>₁ and <em>C</em>₂ :

<em>y</em> (0) = <em>C</em>₁ + <em>C</em>₂ + 2 = 2

→   <em>C</em>₁ + <em>C</em>₂ = 0 … … … [3]

<em>y'(t)</em> = -<em>C</em>₁ exp(-<em>t</em> ) - 3<em>C</em>₂ exp(-3<em>t</em> ) + (-2<em>t</em> ² + 3<em>t</em> + 1) exp(-<em>t </em>) - 3

<em>y'</em> (0) = -<em>C</em>₁ - 3<em>C</em>₂ + 1 - 3 = 2

→   <em>C</em>₁ + 3<em>C</em>₂ = -4 … … … [4]

Solve equations [3] and [4] to get <em>C</em>₁ = 2 and <em>C</em>₂ = -2. Then the particular solution to the initial value problem is

<em>y(t)</em> = -2 exp(-3<em>t</em> ) + (2<em>t</em> ² + <em>t</em> + 2) exp(-<em>t </em>) - 3<em>t</em> + 2

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Elan Coil [88]

Answer:

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Step-by-step explanation:

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Sophie [7]

Answer:

Probability that both cans were regular soda =  \frac{1}{11}

Step-by-step explanation:

Probability = \frac{Desired outcome}{Total possible outcomes}

We are given 12 total number of cans; 4 cans have been accidentally filled with diet soda.

Probability that first can is a regular soda:

Outcome that first can is a regular soda will give us the number of regular soda available which are 4

Using formula of probability

Total possible outcomes are, n(total) = 12

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P(1st can) =  \frac{4}{12} = \frac{1}{3}

Probability that 2nd can is a regular soda:

<em>As we have already taken a can of regular soda from the pack, the total soda in the pack now 11 and the regular soda left are 3.</em>

Total possible outcomes are, n(total) = 11

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P(2nd can) =  \frac{3}{11}


Probability that both cans are regular soda:

P(both) = P(1st can) × P(2nd can)

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             = \frac{1}{11}

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A comparison of the measures of central angles and inscribed angles of circles to their intercepted arcs are listed in the table
geniusboy [140]

A central angle means an angle that is at the center of a circle and whose legs are intersecting the circle.

<h3>What is a central angle?</h3>

Your information is incomplete as the data is not given. Therefore, an overview will be given.

Central angles are shaped by an arc between the two points. When a central angle and the inscribed angle intercept the same arc, then the central angle would be the double of the inscribed angles while the inscribed angle will be half of the central angle.

Learn more about central angle on:

brainly.com/question/19758686

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