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Ganezh [65]
3 years ago
15

What are significant numbers? please explain?

Mathematics
1 answer:
My name is Ann [436]3 years ago
6 0
Non zero digits are always significant. any zeros between two significant digits are significant. a final zero or trailing zero in the decimal portion only are significant.
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What is 15% of 76, rounded to two decimal places?
galben [10]

Answer: 11.4

Step-by-step explanation:

I did part / whole = % / 100

So x / 76 = 15 / 100

76 x 15 = 1140

1140 ÷ 100 = 11.4

6 0
3 years ago
Read 2 more answers
5/8x - 11/2 = 3/4 <br> what does the x equal ?
Verizon [17]

Answer:

x=10

Step-by-step explanation:

5/8x - 11/2 = 3/4

Multiply each side by 8 to clear the fractions

8(5/8x - 11/2 = 3/4)

5x - 44 = 6

Add 44 to each side

5x-44+4= 6+4

5x= 50

Divide by 5

5x/5 = 50/5

x =10

6 0
3 years ago
A carpentry tool has replaceable hexagonal heads. The diagram below shows the replaceable head. *Picture not drawn to scale What
prohojiy [21]

Answer:

Step-by-step explanation:c 15

7 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
3 years ago
Plz help
yKpoI14uk [10]

Answer:

1.  A point you can't move at all, a line you can only move back and forth in the same direction. Yes it is accurate for its characteristics because points and lines have no set definition for them

2. When you are on a point you can not travel at all in any direction while staying on that point. That means you have zero options to travel in. That is why it is said you have zero dimensions.

3. Normal space refers 3 dimensional space that extends beyond the three dimensions of length, width, and height.


4. If you can move backward, forwards, up and down in two different directions it is considered two dimensional. The two dimensional figure is considered a plane. For example, if you took a piece of paper that extended forever in every direction, that in a geometric a sense, is a plane. The piece of paper itself is itself, finite, and you could call the piece of paper a plane segment because it is a segment of an entire plane.

8 0
3 years ago
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