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Ostrovityanka [42]
3 years ago
8

What does the value in front of x represents?

Mathematics
1 answer:
Kipish [7]3 years ago
8 0
The value of in front of x represents how many times that x is being used . For example 2x . x is being used two times .
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Let f(x) represent a function.
STatiana [176]

Answer:

For f(x+\frac{5}{4}), f(x) is translated \frac{5}{4} units left.

For f(x)-\frac{5}{4}, f(x) is translated \frac{5}{4} units down.

Step-by-step explanation:

The transformation of f(x)\rightarrow f(x+C) means that the graph shift to left by C units if C is positive number and shifts to right by C units if C is a negative number.

The transformation of f(x)\rightarrow f(x)+C means that the graph shifts up by C units if C is positive number and shifts down by C units if C is a negative number.

Here, in the transformation of f(x)\rightarrow f(x+\frac{5}{4}),C = \frac{5}{4}>0, so, the function translates left by \frac{5}{4} units.

Similarly, in the transformation of f(x)\rightarrow f(x)-\frac{5}{4},C = -\frac{5}{4}, so, the function translates down by \frac{5}{4} units.

3 0
3 years ago
Kiara made a square baby blanket she decided to add one foot of material to one side then cut 4 inches of material off from the
schepotkina [342]

Answer:

Step-by-step explanation:

960 which is the area u divide by the one foot which is 144 square units to get your answer,6.66666666667

8 0
2 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
Please help with this 8x2+4y÷z when x=12, y=34, and z=2 I don't know this
natita [175]
8x12=96 4x34=136 136+96=232 232/2=116 answer=116
4 0
3 years ago
Ben will purchase four books. The original prices of the books are $29.95, $23.95, $14.95, and $9.95. Ben has four sales options
shutvik [7]

Answer:

Option 2

Step-by-step explanation:

Option 1: $68.85

Option 2: $68.80

Option 3: $70.9

Option 4: $71.81

4 0
2 years ago
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