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sergejj [24]
3 years ago
11

\times 6} 5" alt=" \frac{ - 2 + 3 \times 6} 5" align="absmiddle" class="latex-formula">
I need help please ​
Mathematics
2 answers:
zhuklara [117]3 years ago
7 0

Answer:

Step-by-step explanation:

we do first the multiplication 3*6=18

-2+18=16

16/5 is your answer or 3.2

Alexxandr [17]3 years ago
5 0
-2 + 3 x 6

-2 + 18

16

16/5 or 3.2
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A train traveled 210 miles in 3.5 hours to its first stop. It then continued on traveling another 130 miles in 2.5 hours. What w
valkas [14]
T = 6 [h]
s = 340 [mi]

v=s/t=340/6=56.7 [mi/h]
Answer C.
8 0
3 years ago
If x = 29 but y = x + x - 2x, how can x be represented algebraically through the context of polynomial division
ryzh [129]

Answer:

y will equal 29 + 29 - 2(29) which equals 0

when y = 0, the solution is the x-axis

hope this helps! <3

7 0
1 year ago
GIVING BRAINLIEST Use the inequality 186&lt; -3 (4x - 2).
a_sh-v [17]

Answer:

x < -15

Step-by-step explanation:

I solved the inequality but i don't know if that was what you were asking for lol, if not sorry.

3 0
2 years ago
Help I need this now
eimsori [14]

Answer:

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Step-by-step explanation:

3 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
3 years ago
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