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3 years ago

I'LL GIVE BRAINLIEST !!! FASTERR !

Mathematics

1 answer:

Talja [164]3 years ago

3 0

Answer:

opt. B

Step-by-step explanation:

g=60 because equilateral triangle ang=60

<KMN=120 because of linear pair

<h= 180-(120+29)= 31

g+h= 60+31= 91

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) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra

artcher [175]

Answer:

$y(t)=2e^{3t}(2-5t)$

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2

Using the theorem of the Laplace transform for derivatives, we know that:

$\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)$

Replacing the initial values y(0)=4, y′(0)=2 we obtain

$\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4$

and our differential equation (*) gets transformed in the algebraic equation

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0$

Solving for Y(s) we get

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}$

Now, we brake down the rational expression of Y(s) into partial fractions

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}$

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

and

$\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}$

we know that

$\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}$

and for the first translation property of the inverse Laplace transform

$\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}$

and the solution of our differential equation is

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}$

5 0

3 years ago

Classify the model as exponential growth or exponential decay. Identify the growth or decay factor AND the percent of increase o

sweet-ann [11.9K]

See also brainly.com/question/8975313

The growth factor is 2.

The increase is (2 -1)*100% = 100% per period.

5 0

3 years ago

Write the next three terms of the following sequence 4,5,7,10,14

Law Incorporation [45]

4,5,7,10,14=19,25, 32

8 0

3 years ago

A triangle has sides of LaTeX: 1\frac{1}{6}1 1 6 inches, LaTeX: 1\frac{1}{3}1 1 3 inches, and LaTeX: 1\frac{2}{3}1 2 3 inches. W

Nuetrik [128]

Answer:

4 1/6 inches

Step-by-step explanation:

A triangle has sides of LaTeX: 1\frac{1}{6}1 1 6 inches, LaTeX: 1\frac{1}{3}1 1 3 inches, and LaTeX: 1\frac{2}{3}1 2 3 inches. What is the perimeter of the triangle? Make sure your answer is in simplest form.

Perimeter of Triangle = Side a + Side b + Side c

Perimeter of the Triangle =

( 1 1/6 + 1 1/3 + 1 2/3) inches

Lowest common denominator = 6

= 1 + 1 + 1 +(1/6 + 1/3 + 2/3)

= 3 + (1 + 2 + 4/6)

= 3 + 7/6

= 3 + 1 1/6

= 4 1/6 inches

6 0

3 years ago

A teacher teaches two classes with 8 students each. Each student has a 95% chance of passing their class independent of the othe

dsp73

Answer:

0.4466 = 44.66% probability that, in exactly one of the two classes, all 8 students pass.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they pass, or they do not. The probability of an student passing is independent of other students, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Probability that all students pass in a class:

Class of 8 students, which means that $n = 8$

Each student has a 95% chance of passing their class independent of the other students, which means that $p = 0.95$

This probability is P(X = 8). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{8,8}.(0.95)^{8}.(0.05)^{0} = 0.6634$

Find the probability that, in exactly one of the two classes, all 8 students pass.

Two classes means that $n = 2$

0.6634 probability all students pass in a class, which means that $p = 0.6634$ .

This probability is P(X = 1). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{2,1}.(0.6634)^{1}.(0.3366)^{1} = 0.4466$

0.4466 = 44.66% probability that, in exactly one of the two classes, all 8 students pass.

3 0

3 years ago

I'LL GIVE BRAINLIEST !!! FASTERR !​

I'LL GIVE BRAINLIEST !!! FASTERR !