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k0ka [10]
3 years ago
12

Write as a product 27a^3−(a−b)^3

Mathematics
1 answer:
jarptica [38.1K]3 years ago
6 0

Answer:

  (2a +b)·(13a^2 -5ab +b^2)

Step-by-step explanation:

The factorization of the difference of cubes is a standard form:

  (p -q)^3 = (p -q)(p^2 +pq +q^2)

Here, you have ...

  • p = 3a
  • q = (a-b)

so the factorization is ...

  (3a -(a -b))·((3a)^2 +(3a)(a -b) +(a -b)^2) . . . . substitute for p and q

  = (2a +b)·(9a^2 +3a^2 -3ab +a^2 -2ab +b^2) . . . . simplify a bit

  = (2a +b)·(13a^2 -5ab +b^2) . . . . . . collect terms

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Mr. Johnson sells erasers for $3 each. He sold 96 erasers last week and he sold 204 erasers this week
IgorC [24]

Answer:

288

Step-by-step explanation:

it's pretty simple, if u do 96 * 3, it will be that. yw :)

4 0
3 years ago
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Briainliest Simplify 15√90/√125<br> options<br> 9√2<br> 18√5<br> 18√2/5<br> 27√2/5
nekit [7.7K]

Answer:

9√2

Step-by-step explanation:

simplify the radical by breaking the radicand up into a product of known factors, assuming positive real numbers.

Exact form:9√2

Decimal form: 12.72792206

3 0
3 years ago
Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n
aliya0001 [1]

Answer:

f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...

We first compute the n-th derivative of f(x)=\ln(1+2x), note that

f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\

Now, if we compute the n-th derivative at 0 we get

f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!

and so the Maclaurin series for f(x)=ln(1+2x) is given by

f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2  \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}

3 0
3 years ago
Olaf needs a total of 3 cups of sugar to make 4 cakes. Write and solve an equation to find the number of cups of sugar he needs
irinina [24]

Answer:

3/4 cup

Step-by-step explanation:

7 0
3 years ago
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What is the seventh term of the sequence whose nth term is (n + 1)(n − 2)?
pav-90 [236]
If you plug in 7 for n, you get
(7+1)(7-2).
Simplify that.
(8)(5)
Simplify that and you get the 7th term of the particular sequence is,
40
Hope it helps :D
8 0
3 years ago
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