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Anni [7]
3 years ago
7

What best describes movement of particles in a solid

Mathematics
2 answers:
icang [17]3 years ago
8 0

Answer:

the movement of particles within a solid is normally slower, do to the particles being tighter together than the other classifications.

Fudgin [204]3 years ago
7 0

Answer:The particles in a solid are tightly packed and locked in place. Although we cannot see it or feel it, the particles are moving = vibrating in place.

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The equation 5m=36.25 models a constant rate situation. What is the value of 7m
Bond [772]

Answer:

7m = 50.75

Step-by-step explanation:

5m=36.25

We can solve this for m by dividing each side by 5

5/5m=36.25/5

m = 7.25

We want to find 7m

7(7.25) = 50.75

3 0
3 years ago
Find the 11th term of the following geometric sequence.<br> 1, 3, 9, 27, ...
Natasha_Volkova [10]

Answer:

the answer is 59048

Step-by-step explanation:

as the common ratio is multiplying by 3 and the first term is 1 so from the rule (Tn=ar(power n-1 ) )

so the 11th term is 1*3(power 11-1 ) equal 3 power 10 equal 59048

4 0
3 years ago
I NEED HELP!! pleaseee
Veronika [31]

Answer:

<h3>y = 20</h3><h3 />

Step-by-step explanation:

line RPN = 180 = 4y - 10 + 90 + y

180 - 90 + 10= 5y

100 = 5y

y = 100/5

y = 20

4 0
2 years ago
PLEASE HURRY!!
Novay_Z [31]

Answer:24 i believe

Step-by-step explanation:

7 0
3 years ago
College calculus. I've tried it a couple times but i can't seem to get the right answer
boyakko [2]
The area is the sum of 'n' rectangle areas.
The width of the rectangle is size of interval, (domain size)/n
The height of each rectangle is f(interval) as interval moves along domain.

For this example, domain size = 3-1 = 2
size of interval = 2/n
height varies from f(1) to f(3), increasing by 2/n each time.
f(1+(2/n)i)

Putting this together, the area is the sum of:
\frac{2}{n}*f(1+\frac{2i}{n})

Since you are given the function f(x). Sub input into f(x) to get area in terms of n and i.

f(1+\frac{2i}{n}) = \frac{3(1+\frac{2i}{n})}{(1+\frac{2i}{n})^2 +8} \\  \\ =\frac{3n+6i}{n}*\frac{n^2}{(n+2i)^2 +8n^2} \\  \\ =\frac{3n^2 +6ni}{9n^2+4ni+4i^2}

Finally, the summation is:
\frac{2}{n}*\frac{3n^2 +6ni}{9n^2+4ni+4i^2} = \frac{6n +12i}{9n^2+4ni+4i^2} \\  \\  A =\lim_{n \to \infty} \sum_{i=1}^n \frac{6n +12i}{9n^2+4ni+4i^2}
7 0
3 years ago
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