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alexgriva [62]
2 years ago
14

Help me, iv been having problems with ads going in my phones storage files, what can i do to stop this?

Engineering
2 answers:
Viefleur [7K]2 years ago
7 0

Answer:

so you can do this to avoid paying but you can just try to rest set your phone it sound crazy like you will loose all your contacts but you wont just re- set it. i mean u can try it other wise i dont know :)

Explanation:

hodyreva [135]2 years ago
7 0

Answer:

You can try pluging your phone into your computor and backing up your data onto itunes and factory resetting it then install your data back and see if that works.

Explanation:

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What fraction of the worlds surface is estimated to be arable land?
zzz [600]

I believe the amount of arable land worldwide is 1/3 as a fraction.

6 0
3 years ago
The time delay of a long-distance call can be determined by multiplying a small fixed constant by the number of communication li
aliina [53]

Answer:

We can compute the diameter of the tree T by a pruning procedure, starting at the leaves (external nodes).

  • Remove all leaves of T. Let the remaining tree be T1.
  • Remove all leaves of T1. Let the remaining tree be T2.
  • Repeat the "remove" operation as follows: Remove all leaves of Ti. Let remaining tree be Ti+1.
  • When the remaining tree has only one node or two nodes, stop! Suppose now the remaining tree is Tk.
  • If Tk has only one node, that is the center of T. The diameter of T is 2k.
  • If Tk has two nodes, either can be the center of T. The diameter of T is 2k+1.

Explanation:

We can compute the diameter of the tree T by a pruning procedure, starting at the leaves (external nodes).

  • Remove all leaves of T. Let the remaining tree be T1.
  • Remove all leaves of T1. Let the remaining tree be T2.
  • Repeat the "remove" operation as follows: Remove all leaves of Ti. Let remaining tree be Ti+1.
  • When the remaining tree has only one node or two nodes, stop! Suppose now the remaining tree is Tk.
  • If Tk has only one node, that is the center of T. The diameter of T is 2k.
  • If Tk has two nodes, either can be the center of T. The diameter of T is 2k+1.
4 0
3 years ago
An object whose mass is 251 kg is located at an elevation of 24 m above the surface of the earth. For g-9.78 ms, determine the g
Harlamova29_29 [7]

Answer:

Gravitational Potential =58.914 KJ

Explanation:

We know that

Gravitational Potential Energy = mass\times g\times Height

Given mass = 251 kg

Height= 24 m

g is acceleration due to gravity = 9.78m/s^{2}

Applying values in the equation we get

Gravitational Potential Energy=251X9.78X24 Joules

Gravitational Potential Energy=58914.72 Joules

Gravitational Potential Energy =\frac{58914.72}{1000}KJ= 58.914KJ

4 0
3 years ago
Do heavier cars really use more gasoline? Suppose a car is chosen at random. Let x be the weight of the car (in hundreds of poun
Alex17521 [72]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks necessary data to solve. But I have found the similar question on the internet. So, I will be using the data from that question to solve this question for the sack of concept and understanding.

Data Given:

x = 27 , 44 , 32 , 47, 23 , 40, 34, 52

y = 30, 19,  24,  13 , 29,  19,  21,  14

It is given that,

∑x = 299

∑y = 167

∑x^{2} = 11887

∑y^{2} = 3773

We are asked to verify the above values manually in this question.

So,

1. ∑x = 299

Let's verify it:

∑x = 27 + 44 + 32 + 47 + 23 + 40 + 34 + 52

∑x = 299

Yes, it is equal to the given value. Hence, verified.

2. ∑y = 167

Let's verify it:

∑y = 30 + 19 +  24 + 13 + 29 + 19 +  21 +  14

∑y = 169

No, it is not equal to the given value.

3. ∑x^{2} = 11887

Let's verify it:

For this to find,  first we need to square all the value of x individually and then add them together to verify.

∑x^{2} = 27^{2} + 44^{2} + 32^{2} + 47^{2} + 23^{2} + 40^{2} + 34^{2} + 52^{2}

∑x^{2} = 11,887

Yes, it is equal to the given value. Hence, verified.

4. ∑y^{2} = 3773

Let's verify it:

Again, for this we need to find the squares of all the y values and then add them together to verify it.

∑y^{2} = 30^{2} + 19^{2} +  24^{2} + 13^{2} + 29^{2} + 19^{2} +  21^{2} +  14^{2}

∑y^{2}  = 3,845

No, it is not equal to the given value.

4 0
2 years ago
Water flows through a horizontal plastic pipe with a diameter of 0.15 m at a velocity of 15 cm/s. Determine the pressure drop pe
Sonja [21]

Answer:0.1898 Pa/m

Explanation:

Given data

Diameter of Pipe\left ( D\right )=0.15m

Velocity of water in pipe\left ( V\right )=15cm/s

We know viscosity of water is\left (\mu\right )=8.90\times10^{-4}pa-s

Pressure drop is given by hagen poiseuille equation

\Delta P=\frac{128\mu \L Q}{\pi D^4}

We have asked pressure Drop per unit length i.e.

\frac{\Delta P}{L} =\frac{128\mu \ Q}{\pi D^4}

Substituting Values

\frac{\Delta P}{L}=\frac{128\times8.90\times10^{-4}\times\pi \times\left ( 0.15^{3}\right )}{\pi\times 4 \times\left ( 0.15^{2}\right )}

\frac{\Delta P}{L}=0.1898 Pa/m

4 0
3 years ago
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