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pogonyaev
3 years ago
7

Simplify the expression r^-3 s^5 t^2/r^2 st^-2

Mathematics
1 answer:
Hoochie [10]3 years ago
5 0
R/s^5t^r2st-2/r^3
hope this helps










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32 in standard form is 3.2 X 10^1.

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Step-by-step explanation:

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A spirit shop outside the hockey stadium sells merchandise representing the three school districts that built the arena. the ent
almond37 [142]

Point E is where the statue should be placed.

<h3>What is called a triangle?</h3>
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At what point should the statue be:

The intersection of the doors to the arena, the spirit store, and the parking lot forms a triangle, with Point E in its center. The radius of the circumscribed circle will therefore equal the distance from the statue to each vertex, placing it at the same distance from all three landmarks.

Point E is where the statue should be placed.

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4 0
1 year ago
Use mathematical induction to prove that for each integer n &gt; 4,5" &gt; 2^2n+1 + 100.
Flura [38]

Answer:

The inequality that you have is 5^{n}>2^{2n+1}+100,\,n>4. You can use mathematical induction as follows:

Step-by-step explanation:

For n=5 we have:

5^{5}=3125

2^{(2(5)+1)}+100=2148

Hence, we have that 5^{5}>2^{(2(5)+1)}+100.

Now suppose that the inequality holds for n=k and let's proof that the same holds for n=k+1. In fact,

5^{k+1}=5^{k}\cdot 5>(2^{2k+1}+100)\cdot 5.

Where the last inequality holds by the induction hypothesis.Then,

5^{k+1}>(2^{2k+1}+100)\cdot (4+1)

5^{k+1}>2^{2k+1}\cdot 4+100\cdot 4+2^{2k+1}+100

5^{k+1}>2^{2k+3}+100\cdot 4

5^{k+1}>2^{2(k+1)+1}+100

Then, the inequality is True whenever n>4.

3 0
3 years ago
If y varies directly as x and y=35 when x=7, find y when x=11
sveta [45]
Y=kx
35=7k
35/7=k
k=5

y=5x
y=5×11
y=55
3 0
3 years ago
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