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denis23 [38]
3 years ago
14

40x/64ydivided by 5x/8y=

Mathematics
1 answer:
blsea [12.9K]3 years ago
4 0
<span>40x/64ydivided by 5x/8y would be better written as

</span><span>40x/64y
------------    which in turn would be clearer if written as
</span><span>5x/8y

  40x
--------
   64y
==========
    5x
  ------
    8y

Now invert (5x/8y) and multiply (40x/64y) by your result:

</span>  40x      8 y
-------- * ------    Reducing, we get  8/8, or 1.  The answer is 1.
   64y     5x
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What is the value of the expression 2x² – 5x + 6 when<br> x = -2?
Svet_ta [14]

Answer:

24 is correct answer

Step-by-step explanation:

2 {x}^{2}  - 5x + 6 \\  = 2 {( - 2)}^{2}  - 5 \times ( - 2) + 6 \\  = 2 \times 4 + 10 + 6 \\  = 8 + 10 + 6  \\  = 24

hope it helped you:)

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In the equation (x^2+y)^5, what is the coefficient of the term x^4y^3? what is the coefficient of the same term in the expansion
lys-0071 [83]

\displaystyle&#10;(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^{n-k}y^k

<em>-------------------------------------------------------------</em>


\displaystyle&#10;(x^2+y)^n=\sum_{k=0}^n\binom{n}{k}x^{2n-2k}y^k\\&#10;n=5\\&#10;k=3\\\\\binom{5}{3}=\dfrac{5!}{3!2!}=\dfrac{4\cdor5}{2}=10

<u>It's 10.</u>

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

\displaystyle&#10;(3x^2+y)^n=\sum_{k=0}^n\binom{n}{k}(3x)^{2n-2k}y^k=\sum_{k=0}^n\binom{n}{k}\cdot 3^{2n-2k}\cdot x^{2n-2k}y^k\\\\&#10;n=5\\&#10;k=4\\\\&#10;\binom{5}{3}\cdot3^{2\cdot5-2\cdot4}=10\cdot3^{2}=10\cdot9=90

<u>It's 90</u>

6 0
3 years ago
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