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3 years ago

Simplify to standard form (2x +4) + (-7 – 3b) + (5 – 8x)

Mathematics

1 answer:

docker41 [41]3 years ago

8 0

Answer:

−3b−6x+2

Step-by-step explanation:

2x+4−7−3b+5−8x

=2x+4+−7+−3b+5+−8x

Combine Like Terms:

=2x+4+−7+−3b+5+−8x

=(−3b)+(2x+−8x)+(4+−7+5)

=−3b+−6x+2

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When you cut a pyramid parallel to the base, meaning side to side, what 2D shape is created by the cross section ?

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A batch consists of 12 defective coils and 88 good ones. Find the probability of getting two good coils when two coils are rando

Allushta [10]

Answer:

The probability of getting two good coils when two coils are randomly selected if the first selection is replaced before the second is made is 0.7744

Solution:

Total number of coils = number of good coils + defective coils = 88 + 12 = 100

p(getting two good coils for two selection) = p( getting 2 good coils for first selection ) $\times$ p(getting 2 good coils for second selection)

p(first selection) = p(second selection) = $\frac{\text { number of good coils }}{\text { total number of coils }}$

Hence, p(getting 2 good coil for two selection) = $\frac{88}{100} \times \frac{88}{100} =\bold{0.7744}$

5 0

3 years ago

Solve the equation. 33 = p – 6.71 A. –39.71 B. –26.29 C. 39.71 D. 26.29

jeka57 [31]

Answer

C. 39.71

Explanation

33 = p - 6.71

The first step is to make the like terms to be on the same side.

Add 6.71 on both sides of the eqution

33 + 6.71 = p - 6.71 + 6.71

39.71 = p

∴ p = 39.71

4 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