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Bumek [7]
3 years ago
7

Printer allows for optional settings with a panel of six​on-off switches in a row. how many different settings can she select if

there are no restrictions on the switches​?
Mathematics
1 answer:
Leokris [45]3 years ago
6 0

There are 2×2×2×2×2×2 = 2⁶ = 64 possible settings.

_____

Each switch can be in 2 positions, and for each of those, the other switches can have all possible settings. 1 switch has 2 settings. 2 switches will have 2×2 settings. 3 switches will have 2×2×2 settings.

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The grid below shows figure Q and its image figure Q' after a transformation:
Masteriza [31]
<h3>Answer: choice B) counterclockwise rotation of 90 degrees around the origin</h3>

To go from figure Q to figure Q', we rotate one of two ways

* 270 degrees clockwise

* 90 degrees counterclockwise

Since "270 clockwise" isn't listed, this means "90 counterclockwise" is the only possibility.

4 0
3 years ago
A ½in diameter rod of 5in length is being considered as part of a mechanical linkagein which it can experience a tensile loading
Evgesh-ka [11]

Answer:

a. Maximum Load = Force = 27085.09 N

b. Maximum Energy = 3.440 Joules

Step-by-step explanation:

Given

Rod diameter = ½in = 0.5in

Length = 5in

Young's modulus = 15.5Msi

By applying the 0.2% offset rule,

The maximum load the rod can hold before it gets to breaking point is given as follows by taking the strain as 0.2%

Young Modulus = Stress/Strain ------- Make Stress the Subject of Formula

Stress = Strain * Young Modulus

Stress = 0.2% * 15.5

Stress = 0.002 * 15.5

Stress = 0.031Msi

Calculating the area of the rod

Area = πr² or πd²/4

Area, A = 22/7 * 0.5^4 / 4

A = 22/7 * 0.25 / 4

A = 5.5/28

A = 0.1964in²

The maximum load that the rod would take before it starts to permanently elongate is given by

Force = Stress * Atea

Force = 0.031Msi * 0.1964in²

Force = 31Ksi * 0.1964in²

Force = 6.089Ksi in²

Force = 6.089 * 1000lbf

Force = 6089 lbf

1 lbf = 4.4482N

So, Force = 6089 * 4.4482N

Force = 27085.09 N

b.

Using Strain to Energy Formula

U = V×σ²/2·E

Where V = Volume

V = Length * Area

V = 5 in * 0.1964in²

V = 0.982in³

σ = Stress = 0.031Msi

= 0.031 * 1000Ksi

= 31Ksi

= 31 * 1000psi

= 31000psi

E = Young Modulus = 15.5Msi

= 15.5 * 1000Ksi

= 15.5 * 1000 * 1000psi

= 15500000psi

So,

Energy = 0.982 * 31000²/ ( 2 * 15500000)

Energy = 943,702,000/31000000

Energy = 30.442in³psi

------- Converted to ftlbf

Energy = 2.537 ftlbf

-------- Converted to Joules

Energy = 3.440 Joules

7 0
4 years ago
Line parallel to y=-3x 4 and goes through (-4,6)
Eduardwww [97]
Y=-3x+4
Gradient, m= -3
Parallel lines have equal gradients;
So, equation II,
y=mx+c
y=-3x+c
Replacing for x and y using point (-4, 6)
6=-3(-4)+c
6=12+c
6-12=c
c=-6
y=-3x-6
7 0
3 years ago
What is the slope of the line that passes through the following points: (3,-1) and (-1,2)?
Katyanochek1 [597]

with the equation of the slope for two points given, m=Y2-Y1/ X2-X1, you can obtain the slope , x1= 3, y1 = -1 , x2= -1, y2= 2  

m = 2-(-1)/ -1-3 ⇒ m= 3/-4, m = -3/4, therefore the slope is - 3/4


7 0
3 years ago
The point (17,−9) is translated to the point (6,2). Select the description of the translation.
Novay_Z [31]
Is there any picture to this?
3 0
3 years ago
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