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max2010maxim [7]
3 years ago
14

What is the 25th term of the sequence, where a_1=3 3,7,11,15,19

Mathematics
1 answer:
alekssr [168]3 years ago
7 0

Answer:

99

Step-by-step explanation:

just add lol

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Anni [7]

Answer:

(5,-3)

Step-by-step explanation:

On the x-axis, go 5 points to the right and on the y-axis go 3 points down the the negative 3 and the point N crosses at (5,-3)

7 0
3 years ago
Read 2 more answers
Amy bakes & cakes.
astra-53 [7]

Answer:

48

Step-by-step explanation:

If Amy has 6 cakes and Barry has four times as many as she does then just do 6x4. After doing that you have 24. 24+6=30. Ok so you have 30 so far, Right? And Ceri bakes 12 more then Amy so what's 12+6? 18. 30+18=48.

Your welcome.

4 0
3 years ago
The actual distance between town a and Town B is 64 km. What is the distance on Beth's map? Did you use the graph Or the equatio
saul85 [17]
Use the equation and the map it might help so you can check your answer both ways tell me what the equation is and ill help a little more
4 0
3 years ago
PLZ HELP I BEG
wlad13 [49]

Pythagorean theorem(√82)^2 = 9^2 + x^2

x = 1 cm


15^2 = 9^2 + y^2

y = 12 cm. 

12-1 = 11


Area = 9*11 = 99 cm^2

6 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
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