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Grace [21]
3 years ago
6

3. The positive root of 154 is located between which two consecutive integer values?

Mathematics
2 answers:
telo118 [61]3 years ago
6 0

Answer:

Step-by-step explanation:

12²=144

√144=12

13²=169

√169=13

√154 lies between 12 and 13.

lesantik [10]3 years ago
4 0

Answer:

between 12 and 13

Step-by-step explanation:

answer is 12.4

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natali 33 [55]

Answer:

Step-by-step explanation:

A) either radius or diameter and height

V = πR²h or πD²h/4

B) Radius is half a diameter, R = 4 cm

C) Rather depends on how they are put together.

Placing the rectangular faces touching would make a cylinder

D) V = 3.14(4²)(2) = 100.48  ≈ 100 cm³

E) divide by 2

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F V = ½(3.14)(4²)(2) = 50.24 ≈ 50 cm³

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2 years ago
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Simplify this expression and show all steps:<br> 48m^7/20m^4
Ghella [55]

Answer:

(12/5)m^3

Step-by-step explanation:

It's easier for most of us to write out a fraction / ratio without the " / " symbol:

48m^7

------------

20 m^4

Separate the coefficients from the powers of m:

48     m^7

----- * --------     Now let's reduce each fraction separately:

20      m^4


48/20 = 12/5, and

(m^7) / (m^4) = m^(7-4) = m^3


... so that our final answer is (12/5)m^3.

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2 years ago
What is -2(3x+12y-5-17x-16y+4) simplified?
CaHeK987 [17]
28x + 8y + 2 is the simplified version.
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3 years ago
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A chemist wants to make 100 liters of a 44% acid solution. She has solutions that are 20% acid and 60% acid.
alexgriva [62]

the first solution is 20% acid, and say we'll be using "x" liters, so how many liters of just acid are in it?  well 20% of "x" or namely 0.2x.  Likewise for the 60% acid solution, if we had "y" liters of it, the amount of only acid in it is 0.6y.

\begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{liters of }}{amount}\\ \cline{2-4}&\\ \textit{1st solution}&x&0.20&0.2x\\ \textit{2nd solution}&y&0.60&0.6y\\ \cline{2-4}&\\ mixture&100&0.44&44 \end{array}~\hfill \begin{cases} x+y=100\\\\ 0.2x+0.6y=44 \end{cases}

x+y=100\implies y=100-x~\hfill \stackrel{\textit{substituting on the 2nd equation}}{0.2x+0.6(100-x)=44} \\\\\\ 0.2x+60-0.6x=44\implies -0.4x+60=44\implies -0.4x=-16 \\\\\\ x=\cfrac{-16}{-0.4}\implies \boxed{x=40}~\hfill \boxed{\stackrel{100-40}{y=60}}

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