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zhannawk [14.2K]
3 years ago
14

Give the most appropriate reason for using three significant figures in reporting results of typical engineering calculations.a.

Three significant figures gives better than one-percent accuracy. b. Most of the original data used in engineering calculations do not have accuracy better than one percent. c. Historically slide rules could not handle more than three significant figures. d. Telephone systems designed by engineers have area codes consisting of three figures
Mathematics
1 answer:
Sophie [7]3 years ago
8 0

Answer:

The answer is "Option b".

Step-by-step explanation:

In the calculation of engineering or modeling, it is used to assess its necessary system, that performance, and also used to evaluate the distribution network adjustments, which is used to accommodate the slew of other distribution network adjustments, that's why except "choice b" all were incorrect.

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- Atriangle is 3 cm wider than it is tall. The area is 14 cm^2. Find the height and the base.​
Shkiper50 [21]

Answer:

7 and 4

Step-by-step explanation:

If you set up a rectangle you get an area of 28 cm^2

4x7=28

the base is 7 cm

the height is 4 cm

3 0
3 years ago
Jenny was doing laundry. She folded 10
ipn [44]
10 / p
(10 divided by p)
7 0
3 years ago
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Dentify the terms and like terms in the expression<br> 1/2 s - 4 + 3/4 s + 1/8 - s*3
Aleonysh [2.5K]
Your answer well be-(3*s)+(5/(4*s))-((31/8)) did this help

8 0
4 years ago
Let c be a positive number. A differential equation of the form dy/dt=ky^1+c where k is a positive constant, is called a doomsda
stich3 [128]

Answer:

The doomsday is 146 days

<em></em>

Step-by-step explanation:

Given

\frac{dy}{dt} = ky^{1 +c}

First, we calculate the solution that satisfies the initial solution

Multiply both sides by

\frac{dt}{y^{1+c}}

\frac{dt}{y^{1+c}} * \frac{dy}{dt} = ky^{1 +c} * \frac{dt}{y^{1+c}}

\frac{dy}{y^{1+c}}  = k\ dt

Take integral of both sides

\int \frac{dy}{y^{1+c}}  = \int k\ dt

\int y^{-1-c}\ dy  = \int k\ dt

\int y^{-1-c}\ dy  = k\int\ dt

Integrate

\frac{y^{-1-c+1}}{-1-c+1} = kt+C

-\frac{y^{-c}}{c} = kt+C

To find c; let t= 0

-\frac{y_0^{-c}}{c} = k*0+C

-\frac{y_0^{-c}}{c} = C

C =-\frac{y_0^{-c}}{c}

Substitute C =-\frac{y_0^{-c}}{c} in -\frac{y^{-c}}{c} = kt+C

-\frac{y^{-c}}{c} = kt-\frac{y_0^{-c}}{c}

Multiply through by -c

y^{-c} = -ckt+y_0^{-c}

Take exponents of -c^{-1

y^{-c*-c^{-1}} = [-ckt+y_0^{-c}]^{-c^{-1}

y = [-ckt+y_0^{-c}]^{-c^{-1}

y = [-ckt+y_0^{-c}]^{-\frac{1}{c}}

i.e.

y(t) = [-ckt+y_0^{-c}]^{-\frac{1}{c}}

Next:

t= 3 i.e. 3 months

y_0 = 2 --- initial number of breeds

So, we have:

y(3) = [-ck * 3+2^{-c}]^{-\frac{1}{c}}

-----------------------------------------------------------------------------

We have the growth term to be: ky^{1.01}

This implies that:

ky^{1.01} = ky^{1+c}

By comparison:

1.01 = 1 + c

c = 1.01 - 1 = 0.01

y(3) = 16 --- 16 rabbits after 3 months:

-----------------------------------------------------------------------------

y(3) = [-ck * 3+2^{-c}]^{-\frac{1}{c}}

16 = [-0.01 * 3 * k + 2^{-0.01}]^{\frac{-1}{0.01}}

16 = [-0.03 * k + 2^{-0.01}]^{-100}

16 = [-0.03 k + 0.9931]^{-100}

Take -1/100th root of both sides

16^{-1/100} = -0.03k + 0.9931

0.9727 = -0.03k + 0.9931

0.03k= - 0.9727 + 0.9931

0.03k= 0.0204

k= \frac{0.0204}{0.03}

k= 0.68

Recall that:

-\frac{y^{-c}}{c} = kt+C

This implies that:

\frac{y_0^{-c}}{c} = kT

Make T the subject

T = \frac{y_0^{-c}}{kc}

Substitute: k= 0.68, c = 0.01 and y_0 = 2

T = \frac{2^{-0.01}}{0.68 * 0.01}

T = \frac{2^{-0.01}}{0.0068}

T = \frac{0.9931}{0.0068}

T = 146.04

<em>The doomsday is 146 days</em>

4 0
3 years ago
find two positive even consecutive integers such that a square of the smaller integer is 10 more than the larger integer
nalin [4]

Answer:

the two positive consecutive integers are 4 and 6.

Step-by-step explanation:

Let the smaller integer be s; then s^2 = (s + 2) + 10.

Simplifying, s^2 - s - 2 - 10 = 0, or

s^2 - s - 12 = 0.

Solve this by factoring:  (s - 4)(s + 3) = 0.

Then s = 4 and s = -3.

If the first even integer is 4, the next is 6.  We omit s = -3 because it's not even.

The smaller integer is 4.  Does this satisfy the equation s^2 = (s + 2) + 10?

4^2 = (4 + 2) + 10  True or False?

16 = 6 + 10 = 16.

True.

So the two positive consecutive integers are 4 and 6.

6 0
3 years ago
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