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Kryger [21]
3 years ago
9

1.The HCF of 15 and 20?2.5!=??...????​

Engineering
2 answers:
nydimaria [60]3 years ago
7 0

Answer:

the h.c.f of 15and20 is 5

kkurt [141]3 years ago
7 0

Answer:

The answer is 5.

Explanation:

15:3×5

20:2×2×5

In both 15 and 20, '5' is in common. So, the numbers which are in common are called HCF ( Highest Common Factor ).

So, the HCF (well known as GCF ) of 15 and 20 will be 5.

I hope this helped you. If this helped you please mark this answer as brainliest because, I want to go in virtuoso rank and I still need 3 more points.

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Please write the following code in Python 3. Also please show all output(s) and share your code.
maksim [4K]

Answer:

sum2 = 0

counter = 0

lst = [65, 78, 21, 33]

while counter < len(lst):

   sum2 = sum2 + lst[counter]

   counter += 1

Explanation:

The counter variable is initialized to control the while loop and access the numbers in <em>lst</em>

While there are numbers in the <em>lst</em>,  loop through <em>lst</em>

Add the numbers in <em>lst</em> to the sum2

Increment <em>counter</em> by 1 after each iteration

6 0
3 years ago
A 3-in-thick slab is 10 in wide and 15 ft long. The thickness of the slab is reduced by 20% and width increases by 3% in a hot-r
xxMikexx [17]

Answer: l = 2142.8575 ft

v = 193.99 ft/min.

Explanation:

Given data:

Thickness of the slab = 3in

Length of the slab = 15ft

Width of the slab = 10in

Speed of the slab = 40ft/min

Solution:

a. After three phase

three phase = (0.2)(0.2)(0.2)(3.0)

= 0.024in.

wf = (1.03)(1.03)(1.03)(10.0)

= 10.927 in.

Using constant volume formula

= (3.0)(10.0)(15 x 15) = (0.024)(10.927)Lf

Lf = (3.0)(10.0)(15 x 15)/(0.024)(10.927)

= 6750 /0.2625

= 25714.28in = 2142.8575 ft

b.

vf = (0.2 x 0.2 x 3.0)(1.03 x 1.03 x 10.0)(40)/(0.024)(10.927)

= (0.12)(424.36)/0.2625

= 50.9232/0.2625

= 193.99 ft/min.

4 0
3 years ago
What is the core domain for Accenture’s Multi-party Systems practice?
12345 [234]

Answer:

a

Explanation:

digital identity is the answer

6 0
3 years ago
III. During January, at a location in Alaska winds at −27°C can be observed. However, several meters below ground the temperatur
Naya [18.7K]

Answer:

Not possible.

Explanation:

According to second law of thermodynamics, the maximum efficiency any heat engine could achieve is Carnot Efficiency η defined by:

\eta=1-\frac{T_{cold}}{T_{hot}}

Where

T_{hot} and T_{cold} are temperature (in Kelvin) of heat source and heatsink respectively

In our case (I will be using K = 273+°C) :

\eta=1-\frac{-27+273}{14+273}\\=0.1428

In percentage, this is 14.28% efficiency, which is the <em>maximum</em> theoretical efficiency <em>any</em> heat engine could have while working between -27 and 14 °C temperature. Any claim of more efficient heat engine between these 2 temperature are violates the second law of thermodynamics. Therefore, the claim must be false.

6 0
3 years ago
CAR people need some help fr <br><br>which cars should be my first super car pls
hammer [34]

Answer:

nah none of those i would get a tesla, mclaren, corvette, or even a toyota camery or tundra!

Explanation:

my dads is really big into cars and stuuf so i learn a little lol! hope it helps! thanks for the points by the way!

7 0
3 years ago
Read 2 more answers
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