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3 years ago

Math question down below

Mathematics

1 answer:

sergeinik [125]3 years ago

7 0

The absolute value symbol (that being when a number has the 2 lines on the outside |-x| simply makes whatever is inside of it positive after it has been solved. Therefore, the easiest way to think about it is without the absolute value signs at all and just assume every number is positive. When we do this, we can see that the correct answer is C. |-3.5|, |4.2|, |-9|.

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As a salesperson, Jonathan is paid $50 per week plus 3% of the total amount he sells. This week, he wants to earn at least $100.

Murljashka [212]

Answer:

50+0.03x $\geq 100$

Step-by-step explanation:

All you need to do is look at the numbers given in the word problem and convert them to algebraic terms.

5 0

3 years ago

Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$

$L^{-1}[\dfrac{e^{-3s}}{s+1}]$

$L^{-1}[\dfrac{1}{s+1}] = e^{-t} = f(t) \ then \ by \ second \ shifting \ theorem;$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$= e^{-t-3} \left \{ {{1 \ \ \ \ \ t>3} \atop {0 \ \ \ \ \ t$

= $e^{-(t-3)} u (t-3)$

Recall that:

$y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$

Then

$y(t) = 7 -7e^{-t} +e^{-(t-3)} u (t-3)$

$y(t) = 7 -7e^{-t} +e^{-t} e^{-3} u (t-3)$

$\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

3 0

2 years ago

Eric made a drawing of his rectangular bedroom with the scale 1 inch=3 feet. The drawing was 5 inches long 4 inches wide. What w

kolbaska11 [484]

Answer:

180 square feet

Step-by-step explanation:

We know that the rectangular drawing has:

5 inches long (L)
4 inches wide (W)

And we also know that:

1 inch = 3 feet

We will find how many feet long and wide is the actual room, by multiplying:

Long: 5 inches * 3 feet/inch = 15 feet

Wide: 4 inches * 3 feet/ inch = 12 feet

Now we want to know the area of the rectangle, and the area of the rectangle is:

A = L*W

A = 15 feet * 12 feet

A = 180 square feet

3 0

3 years ago

to multiply a number by 12 carter likes to multiply the number by 10 and then multiply it by 2 and add the product. here is a pi

Elanso [62]

Poop

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3 years ago

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The axis of symmetry for a function in the form f(x) = x2 + 4x - 5 is x = -2. What are the coordinates of the vertex of the

umka2103 [35]

Answer:

(- 2, - 9 )

Step-by-step explanation:

The axis of symmetry passes through the vertex of the parabola.

Thus the x- coordinate of the vertex is x = - 2

To find y substitute x = - 2 into f(x)

f(- 2) = (- 2)² + 4(- 2) - 5 = 4 - 8 - 5 = - 9

Vertex = (- 2, - 9 )

8 0

2 years ago

Read 2 more answers