Simplify (8x2 − 1 2x3) − (7x3 − 3x2 1). −5x3 11x2 − 2 5x3 − 11x2 2 x3 2x2 x3 x3 − 2x2 − x3.

Mathematics

2 answers:

Harrizon [31]2 years ago

8 0

Answer:

-5x^3 +11x^2-2

(8x^2 − 1 + 2x^3) − (7x^3 − 3x^2 + 1)Distribute the minus sign(8x^2 − 1 + 2x^3) − 7x^3 + 3x^2 - 1I like to line them up vertically from largest to smallest2x^3 + 8x^2 − 1− 7x^3 + 3x^2 - 1------------------------------5x^3 +11x^2-2

correct me if im wrong

Komok [63]2 years ago

7 0

Answer:

-5x^3 +11x^2-2

Step-by-step explanation:

(8x^2 − 1 + 2x^3) − (7x^3 − 3x^2 + 1)

Distribute the minus sign

(8x^2 − 1 + 2x^3) − 7x^3 + 3x^2 - 1

I like to line them up vertically from largest to smallest

2x^3 + 8x^2 − 1

− 7x^3 + 3x^2 - 1

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

-5x^3 +11x^2-2

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A store has apples on sale for $14.00 for 4 pounds. If an apple is approximately 5 ounces, how many apples can you buy for $84.0

adelina 88 [10]

For blank 1:
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For blank 2:
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4 0

3 years ago

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If 3t-7=5t, then 6t =

Natalija [7]

The answer is: " -21 " .
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We are asked to solve for: "6t" ; which is: "6 * t" .
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Given: 3t − 7 = 5t ; Solve for "t" ; then solve for "6t" ;

     3t − 7 = 5t ;   Subtract "3t" from each side of the equation;
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    3t − 7 − 3t = 5t − 3t ;

to get: -7 = 2t ;

Divide EACH side of the equation by "2" ; to isolate "t" on one side of the equation; and to solve for "t" ;

-7/2 = 2t / 2 ;

-3.5 = t ; ↔ t = - 3.5 ;
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Now, solve for "6t" ;

6 t = 6*(-3.5) = - 21 .   The answer is: " -21 " .
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3 years ago

Please calculate this limit <br>please help me

Tasya [4]

Answer:

We want to find:

$\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n}$

Here we can use Stirling's approximation, which says that for large values of n, we get:

$n! = \sqrt{2*\pi*n} *(\frac{n}{e} )^n$

Because here we are taking the limit when n tends to infinity, we can use this approximation.

Then we get.

$\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n} = \lim_{n \to \infty} \frac{\sqrt[n]{\sqrt{2*\pi*n} *(\frac{n}{e} )^n} }{n} = \lim_{n \to \infty} \frac{n}{e*n} *\sqrt[2*n]{2*\pi*n}$