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vladimir1956 [14]
3 years ago
11

Help homework is due today. please

Mathematics
1 answer:
Oxana [17]3 years ago
8 0
First solve the equation:

5(3x-7)=20 => 15x - 35 = 20 => 15x = 55 => x=11/3

Fill in this solution in 6x-8:

6*11/3 - 8 = 22 - 8 = 14

Second question: apply the distributive property (ie,. both terms get multiplied by 4/5):

4/5(20x - 10) = 80/5x - 40/5 = 16x - 8
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350 cm^2 (A for connections academy)
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3 years ago
Morgan is making accessories for the soccer team. She uses 822.48 inches of fabric on headbands for 43 players and 3 coaches. Sh
emmasim [6.3K]

Answer:

headband 17.88

Wristband 9.31

Step-by-step explanation:

headband: 822.48/(43+3)

Wristband: 400.33/43

6 0
3 years ago
Jack got the expression 7x + 1 and then wrote his answer as 1 + 7x. Is his answer an equivalent expression? How do u know? Expla
nadezda [96]
Yes, it is an equivalent expression because even though you move two pieces of an equation it will still equal the same thing, unless there are parentheses involved.
6 0
3 years ago
Read 2 more answers
) 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
Can someone please explain?
GREYUIT [131]

The function y = x, called the "identity" or "ramp" function, is a basic function that you'll need to add to your math vocabulary. Since the " x " here seems to have no exponent, fix that by thinking "x^1," or "x to the first power," or "y=x is a linear function."

The graph of y=x always goes thru the origin. It begins in the 3rd quadrant and ends in the 1st, and appears as a straight line with slope of m = rise / run = 1/1 = 1.

8 0
3 years ago
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