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3 years ago

If M is the Midpoint of AB, then solve for x find the unknown measures the given segments.

Mathematics

1 answer:

Leto [7]3 years ago

8 0

Answer:

If M is midpoint of AB, then

AM=MB

4x-5= 2x+11

2x=16

x=8

AM= 4x-5

=4(8)-5

=32-5

=27

BM=2x+11

=2(8)+11

=16+11

=27

Hope it helps:-)

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3 years ago

Solve for x: x/8 - x/9 = 1

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Step-by-step explanation:

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2 years ago

If f(x) = e^X, and W(f, g) = 3e^x, find g(x).

trasher [3.6K]

Answer:

g(x)=3

Step-by-step explanation:

Let's find the answer.

W(f,g)=3e^x which can be written as:

W(f,g)=(3)*(e^x), notice that:

(e^x)=f(x) so:

W(f,g)=3*f(x), establishing:

W(f,g)=g(x)*f(x) then:

g(x)=3

In conclusion, g(x)=3.

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3 years ago

What is 8 3/5 + 4 4/5?

dangina [55]

Answer:

67/5 = 13 2/5

Step-by-step explanation:

Step 1: Define/explain.

An easier way to solve this is by changing the mixed fractions to improper fractions.

To do this, multiply the whole number by the denominator, then add the product to the numerator; the denominator remains the same.

Mixed fraction - a fraction with a whole number.

Improper fraction - a fraction with a numerator larger than the denominator.

Step 2: Solve.

$8\frac{3}{5} =\frac{43}{5}$

$4\frac{4}{5} =\frac{24}{5}$

From here, add as usual.

$\frac{43}{5} +\frac{24}{5} =\frac{67}{5}$

Step 3: Conclude.

You can change the improper fraction to a mixed fraction if you'd like.

To do this, divide the numerator by the denominator.

The amount of times the numerator evenly goes into the denominator is the whole number.

The amount of remaining numbers in the denominator.

The numerator remains the same.

$\frac{67}{5} =13\frac{2}{5}$

I, therefore, believe the answer to this is 13 2/5.

5 0

3 years ago

) 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