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Flura [38]
3 years ago
13

Simplify 5+{6[12+(1−3)2]−4}

Mathematics
2 answers:
natulia [17]3 years ago
6 0
We Need To Use Pemdas.
First, 1-3 = -2.
<span>5+{6[12+(-2)2]−4}
Next, -2 * 2 = -4.
</span><span>5+{6[12+(-4)]−4}
Next, 12 + (-4) = 8.
</span><span>5+{6[8]−4}
Next, 6 * 8 = 48.
</span><span>5+{48−4}
Next, 48 - 4 = 44
</span><span>5+44 = 49
So, Your Answer Is 49.</span>
CaHeK987 [17]3 years ago
4 0
The answer is simply 9 :)
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How much would $600 invested at 8% interest compounded continuously be worth after 3 years? Round your answer to the nearest cen
eduard

Answer:

yes

Step-by-step explanation:

You just do it

5 0
3 years ago
What is the length of the diagonal of a poster board with dimensions 22 inches by 28 inches? Round to the nearest tenth.
devlian [24]

Answer:

The length of the diagonal of a poster board is 35.6\ in

Step-by-step explanation:

Let

x----> the length of the diagonal of a poster board

we know that

Applying the Pythagoras Theorem

x^{2}=22^{2}+28^{2} \\ \\x^{2}=1,268\\ \\x=35.6\ in

4 0
3 years ago
Read 2 more answers
The limit as h approaches 0 of (e^(2+h)-e^2)/h is ?
Whitepunk [10]
<span>If you plug in 0, you get the indeterminate form 0/0. You can, therefore, apply L'Hopital's Rule to get the limit as h approaches 0 of e^(2+h),
which is just e^2.
</span><span><span><span>[e^(<span>2+h) </span></span>− <span>e^2]/</span></span>h </span>= [<span><span><span>e^2</span>(<span>e^h</span>−1)]/</span>h

</span><span>so in the limit, as h goes to 0, you'll notice that the numerator and denominator each go to zero (e^h goes to 1, and so e^h-1 goes to zero). This means the form is 'indeterminate' (here, 0/0), so we may use L'Hoptial's rule:
</span><span>
=<span>e^2</span></span>
3 0
3 years ago
Consider the game of independently throwing three fair six-sided dice. There are six combi- nations in which the three resulting
Murrr4er [49]

Answer:

See explanation below.

Step-by-step explanation:

1) First let's take a look at the combinations that sum up 10:

  1. 1+3+ 6,
  2. 1+ 4+ 5,
  3. 2+2+6,
  4. 2+3+5,
  5. 2 + 4 + 4,
  6. 3+3+4

Notice that when we have 3 different numbers on the dice, we can permute them in 6 different ways. For example: Let's take 1 + 3 + 6, we can get this sum with these permutations:

1 + 3 + 6, 1 + 6 + 3, 3 + 6 + 1, 3 + 1 + 6, 6 + 1 + 3, 6 + 3 + 1.

And when we have two different numbers on the dice, we can permute them in 3 different ways:

2 + 2 + 6, 2 + 6 +2, 6 + 2 + 2.

So now we're going to write down the 6 combinations that sum up 10 but we're going to write down how many permutations of them we get:

  1. 1+3+ 6 : 6 permutations
  2. 1+ 4+ 5 : 6 permutations
  3. 2+2+6: 3 permutations
  4. 2+3+5: 6 permutations
  5. 2 + 4 + 4: 3 permutations
  6. 3+3+4: 3 permutations

Total of permutations: 6 + 6 + 3 + 6 + 3 + 3 =27.

Thus we have 27 different ways of getting a sum of 10.

2) Now we're going to take a look at the combinations that sum up 9 and we're going to proceed in a similar way:

  1. 1 + 2 + 6: 6 permutations
  2. 1+3+5: 6 permutations
  3. 1+4+4: 3 permutations
  4. 2+ 3+ 4: 6 permutations
  5. 2+2 +5: 3 permutations
  6. 3+3+3: 1 permutation.

Total of permutations: 6 + 6 + 3 + 6 +3 + 1 = 25.

Thus we have 25 different ways of getting a sum of 10

And we can conclude that the probability of getting a total of 10 is larger than the probability to get a total of 9.

5 0
4 years ago
The 1985 Mexico City earthquake measured a magnitude 8.0 on the Richter scale. The Richter scale uses the function M (I) = log (
Travka [436]

Answer:

The correct option is C: I = 10⁸I₀.   

Step-by-step explanation:

The function of the Richter scale is:

M_{(I)} = log(\frac{I}{I_{0}})

Since the magnitude of the earthquake is 8.0, the relation between the intensity, I, with the intensity of the threshold quake, Io, is:

8.0 = log(\frac{I}{I_{0}})

10^{8} = \frac{I}{I_{0}}

I = 10^{8}I_{0}

Therefore, the correct option is C: I = 10⁸I₀.                

I hope it helps you!                    

7 0
3 years ago
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