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kirill [66]
3 years ago
13

Write 9 decimals with one decimal place that when rounded to the nearest one round to 7

Mathematics
2 answers:
FinnZ [79.3K]3 years ago
8 0
7.030124314  they all just have to be less then 5
balu736 [363]3 years ago
8 0
7.030124314 have to be less than 5
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The quotient of n and 12 subtracted from 100?
Korvikt [17]

Answer:

100 - 12/n

Step-by-step explanation:

quotient means division so

12/n

and subtracted from means say x - 12/n so

100 - 12/n

If you're looking to evaluate it then it would be

4(25n - 3) / n

5 0
2 years ago
The second of 2 numbers is 5 more than twice the first. Their sum is 29. Find the numbers. (With steps)
hodyreva [135]
So the equation is x + 2x + 5= 29
Here are the steps            -5    -5
                               -----------------------
                                        3x   = 24
      Divide both sides by 3 so you get
                                          x = 8
Then look back at the question, the second number has to be twice that number plus 5
                 So the 1st number is 8 The second # is 2(8)+5= 21

8 0
3 years ago
The product of two rational numbers is (-28/81). If one of them is (-2/3), the find the other.
KatRina [158]

Answer:

The other rational number is \dfrac{14}{27}.

Step-by-step explanation:

First rational number is \dfrac{-2}{3}

Let other rational number is \dfrac{x}{y}

The product of two rational numbers is \dfrac{-28}{81}

According to question,

\dfrac{x}{y}\times \dfrac{-2}{3}=\dfrac{-28}{81}

Multiplying both sides by (-3/2) such that,

\dfrac{x}{y}\times \dfrac{-2}{3}\times \dfrac{-3}{2}=\dfrac{-28}{81}\times \dfrac{-3}{2}\\\\\dfrac{x}{y}=\dfrac{14}{27}

So, the other rational number is \dfrac{14}{27}.

6 0
3 years ago
Read 2 more answers
Please help me Algebra 1
Oduvanchick [21]

Answer:

a.  \sqrt{x^n} = x^\frac{n}{2}

b. \sqrt{x^n}  = x^\frac{(n-1)}{2}\sqrt{x}

Step-by-step explanation:

a) When <em>n </em>is even, then it is divisible by 2. Because of this, you can write:

  • \sqrt{x^n} = (x^n)^\frac{1}{2}
  • \sqrt{x^n} = x^\frac{n}{2}

b) When <em>n </em>is odd, then <em>n - 1 </em> is even. This would make it divisible by 2, and there would be a remainder of 1, so we can write:

  • \sqrt{x^n} = (x^n^-^1^+^1) ^\frac{1}{2}
  • \sqrt{x^n} = (x^n^-^1 × x)^\frac{1}{2}
  • \sqrt{x^n} = x^\frac{(n-1)}{2} × x^\frac{1}{2}
  • \sqrt{x^n}  = x^\frac{(n-1)}{2}\sqrt{x}
8 0
2 years ago
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On my paper I got a 96 but I want a 100 so I am doing corrections on one problem and I need help please. I tried 5 but that wasn
Scorpion4ik [409]
16 students out of 20
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2 years ago
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