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bixtya [17]
3 years ago
15

Try the following example. Remember to use the hints if you need them. Solve for x. 2.9×10−7=(x)(x)2(4.6)(4.6)2(3)2 Choose the c

orrect answer. View Available Hint(s) Choose the correct answer. 4.0×10−6 2.3 2.3×10−2 I don’t know. I am unable to obtain any of the answers but I do not want to give up.
Mathematics
1 answer:
galben [10]3 years ago
7 0

Answer:

2.3\times 10^{-2}

Step-by-step explanation:

Given equation,

2.9\times 10^{-7}=\frac{(x)(x)^2(4.6)}{(4.6)^2(3)^2}

2.9\times 10^{-7}=\frac{x^3}{9\times 4.6}

(\because a^m = \frac{1}{a^{-m}})

By cross multiplication,

2.9\times 9\times 4.6\times 10^{-7}=x^3

120.06\times 10^{-7}=x^3

\implies x = (120.06\times 10^{-7})^\frac{1}{3}

=0.0228980999294

\approx 0.023

=2.3\times 10^{-2}

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Step-by-step explanation:

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3 years ago
Find: (4x2y3 + 2xy2 – 2y) – (–7x2y3 + 6xy2 – 2y) Place the correct coefficients in the difference.
DanielleElmas [232]

Answer: Our required answer will be

11x^2y^3-4xy^2

Step-by-step explanation:

Since we have given that

(4x^y^3+2xy^2-2y)-(-7x^2y^3+6xy^2-2y)

We need to get the difference:

1) We first open the second bracket and make appropriate change in signs :

4x^y^3+2xy^2-2y+7x^2y^3-6xy^2+2y

2) Gather the like terms together :

4x^2y^3+2xy^2-2y+7x^2y^3-6xy^2+2y\\\\=4x^2y^3+7x^2y^3+2xy^2-6xy^2-2y+2y\\\\=11x^2y^3-4xy^2

Hence, our required answer will be

11x^2y^3-4xy^2

5 0
3 years ago
Read 2 more answers
Which of the following methods of timekeeping is the least precise?
katen-ka-za [31]
I think C would be your answer
8 0
3 years ago
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6. A tank contains 100-gallon of pure water. At time t = 0, a solution containing 2 lb of salt per gallon flows into the tank at
SOVA2 [1]

Answer:

The answer is below

Step-by-step explanation:

Let Q represent the amount of salt in the tank at time t.

\frac{dQ}{dt}= flow\ in - flow\ out\\ \\flow\ in=3\ gal/min*2\ lb/gal=6\ lb/min\\\\net\ gain\ in\ tank\ volume=3-4=-1, henceflow\ out= \frac{4Q}{100-t} \\\\\frac{dQ}{dt}= 6-\frac{4Q}{100-t} \\\\\frac{dQ}{dt}+ \frac{4Q}{100-t}=6\\\\The \ integrating\ factor\ is:\\\\IF=e^{\int\limits {\frac{4}{100-t} } \, dt }=e^{-4\int\limits {\frac{-1}{100-t}}=e^{-4ln(100-t)}=(100-t)^{-4}}\\\\Multiplying\ through  \ by\ IF: \\\\(100-t)^{-4}\frac{dQ}{dt}+ (100-t)^{-4}\frac{4Q}{100-t}=6(100-t)^{-4}\\\\

Integrating:\\\\A(100-t)^{-4}=-2(100-t)^{-3}+c\\\\A=-2(100-t)+\frac{c}{(100-t)^{-4}} \\\\at, t=0,A=0\\\\0=-2(100-0)+\frac{c}{(100-0)^{-4}}\\\\c=0.02\\\\A=-2(100-t)+\frac{0.02}{(100-t)^{-4}}

7 0
3 years ago
What is the quotient (x3 + 3x2 + 5x + 3) ÷ (x + 1)?
Sergeeva-Olga [200]
X^3 +3x^2 +5x +3 / x+1 = x^2 +2x +3
x^3 +x^2
------------
 0   2x^2 +5x
      2x^2 +2x
   ----------------
        0      3x +3
                3x +3
               ---------
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so the 3rd choice is right sure 
5 0
4 years ago
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