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drek231 [11]
3 years ago
8

What is the answer to this math problem?

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
8 0

Answer:

The answer is 243 I believe

Step-by-step explanation:

Multiply 27 times 6, once you get the product of that, multimply that by 3. Once you have done all of that multiply that by 1/2 or 0.5

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professor190 [17]
There are 10 possible numbers. 111 is one of the numbers. The probability is 1/10 or 0.1.
There are two numbers that are smaller than 107. The probability is 2/10 or 0.2.

**Side note**: Computers actually are not random, they are pseudorandom. They are very close to random, but not quite. However this will not affect the answer.
4 0
3 years ago
Read 2 more answers
Help me plzzzz<br><br> (x+1)^2 -3(x+2)<br><br> If you could give an explanation that would be great!
Margarita [4]

Answer:

x^2-x-5

Step-by-step explanation:

1. expand the square

(+1)(+1)−3(+2)

2. distribute

(+1)+1(+1)−3(+2)

3. distribute

^2++1(+1)−3(+2)

4. multiply by 1

^2+++1−3(+2)

5. combine like terms

^2+2+1−3(+2)

6. distribute

^2+2+1−3−6

7. subtract the numbers

^2+2−5−3

8. combine like terms

^2−−5

9. gg you are done

^2−−5

4 0
3 years ago
Water is being drained out of a swimming pool at a constant rate of 780 gallons per hour. The swimming pool initially contained
nata0808 [166]

Answer:

Step-by-step explanation:

5 0
3 years ago
1. If a, b, c, d, and e are whole numbers and a(b(c
mars1129 [50]

Answer:

a CANNOT be even ⇒ answer A

Step-by-step explanation:

* Lets revise the rules of even and odd numbers

- Even numbers any number its unit digit is (0 , 2 , 4 , 6 , 8)

- Odd numbers any number its unit digit is (1 , 3 , 5 , 7 , 9)

# even + even = even ⇒ 2 + 4 = 6  

# odd + odd = even ⇒ 1 + 3 = 4  

# odd + even = odd ⇒ 1 + 2 = 3  

# even × even = even ⇒ 2 × 4 = 8

# odd × odd = odd  

⇒ 3 × 5 = 15

# odd × even = even  ⇒ 5 × 6 = 30

∵ a[b(c + d) + e] = odd

∵ odd × odd = odd  

∴ a must be odd and [b(c + d) + e] must be odd

∵ odd + even = odd

∵ odd × even = even

# Case 1

∴ b(c + d) must be odd  if e even

∵ b(c + d) is odd

∴ b must be odd and (c + d) must be odd

∵ c + d must be odd

∵ odd + even = odd

∴ c or d can be even

- We now now that e , c and d can be even

# case 2

∴ b(c + d) must be even  if e odd

∵ b(c + d) is even

∵ even × even = even

∴ b and (c + d) both can be even

∵ c + d can be even

∴ c or d can be even or odd

- We now now that e , c , d and b can be even

∴ Only a can not be even

* a CANNOT be even

5 0
3 years ago
The number of <img src="https://tex.z-dn.net/?f=x" id="TexFormula1" title="x" alt="x" align="absmiddle" class="latex-formula"> s
Alex777 [14]

Answer:

D. 4

Step-by-step explanation:

Without actually solving the equation, recall that for a=|b|, there are two cases:

\begin{cases}a=b, \\a=-b\end{cases}

In the given equation |x-2|^{10x^2-1}=|x-2|^{3x}, there are two pairs of absolute value symbols.

Since each has two cases, there must be a total of 2\cdot 2=\boxed{4} different equations created.

All four cases are:

\begin{cases}(x-2)^{10x^2-1}=(x-2)^{3x},\\(-x+2)^{10x^2-1}=(x-2)^{3x},\\(x-2)^{10x^2-1}=(-x+2)^{3x},\\(-x+2)^{10x^2-1}=(-x+2)^{3x}\end{cases}

Exponents differ, hence clearly there are four possible solutions to this equation.

You can solve for all four values of x by taking the log of both sides and using a bit of algebra to verify you have four solutions.

5 0
3 years ago
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