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Natalka [10]
3 years ago
12

Simply (3x³ + 2x-3) - (4x³ - x² + x)​

Mathematics
1 answer:
Vinvika [58]3 years ago
6 0

Answer:

\left(3x^3+2x-3\right)-\left(4x^3-x^2+x\right)=-x^3+x^2+x-3\\\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\\3x^3+2x-3-\left(4x^3-x^2+x\right)\\3x^3+2x-3-4x^3+x^2-x\\-x^3+x^2+x-3

I hope this is good enough:

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The prime factorization of a number n can be written as n=pr whose p and r are distinct primes. how many factors does n have not
GenaCL600 [577]
It has 3 factors, p, r, and n.
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3 years ago
Karmen returned a bicycle to Earl's Bike Shop. The sales receipt showed a total paid price of $211.86, including the 7% sales ta
MakcuM [25]

Answer:

$198

Step-by-step explanation:

198x.07=13.86

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7 0
4 years ago
The diagrams show a right angled triangle and a rectangle.
Mrrafil [7]

Answer:

hope it helps you......................

8 0
3 years ago
Read 2 more answers
Worth 12 Points, i spent 24
rewona [7]

Answer:

  \displaystyle m^{\frac{35}{12}}n^{-\frac{28}{15}}=\frac{m^{\frac{35}{12}}}{n^{\frac{28}{15}}}

Step-by-step explanation:

The relevant rule of exponents is ...

  (a^b·c^d)^e = a^(be)·c^(de)

Then ...

  (m^(5/4)·n^(-4/5))^(7/3) = m^(5/4·7/3)·n^(-4/5·7/3)

  = m^(35/12)·n^(-28/15)

__

Since you want positive rational exponents, you can write this as ...

  = m^(35/12)/n^(28/15)

6 0
3 years ago
7 cos x+1=6 secx solve for x
jonny [76]
7cos(x) + 1 = 6sec(x)
7cos(x) + 1 = 6/cos(x)
7cos^(x) + cos(x) = 6
7cos^(x) + cos(x) - 6 = 0
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cos(x) = 6/7 , x = arccos(6/7) and
cos(x) = -1, x = 180
6 0
3 years ago
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