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SOVA2 [1]
1 year ago
6

t="p^{3} at \\\\ p = 2" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
WITCHER [35]1 year ago
6 0
<h3>Answer:  8</h3>

Reason:

Replace p with 2 and evaluate.

p^3 = 2^3 = 2*2*2 = 8

Think of a cube that is 2 units along each side. It's volume is 2*2*2 = 8 cubic units. Repeated multiplication can be shortened to using exponents.

Another example: 7^4 = 7*7*7*7

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Javier is buying food in lunch line.the tray of salad plates I'd 3/8tull.the tray of fruit plate 3/4 full.ehich tray is more ful
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The fruit plate is more full
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Find the zeros of the polynomial function.
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◆ QUADRATIC EQUATIONS ◆

given \: equation \: is \: , \\  \\  {x}^{2}  - 12x + 20 = 0 \\  \\  using \: quadratic \: formula \: , \\  \\ x =  \frac{ - b +  \sqrt{ {b}^{2} - 4ac } }{2a}  \:  \:  \:  \:  =   \frac{ - ( - 12) +  \sqrt{ { (- 12)}^{2} - 4(1)(20) } }{2(1)} \\  \\ x =  \frac{ - b +  \sqrt{ {b}^{2} - 4ac } }{2a}  \:  \:  \:  \:  =   \frac{ - ( - 12)  -   \sqrt{ { (- 12)}^{2} - 4(1)(20) } }{2(1)} \\  \\ x =  \frac{12 + 8}{2}  \:  \:  \:  \:  \: or \:  \:  \:  \: x =  \frac{12 - 8}{2}  \\  \\ x = 10 \:  \:  \:  \:  \: or \:  \:  \:  \:  \: x = 2 \:  \:  \: ans.
8 0
3 years ago
(08.01, 8.02 MC) An expression is shown below: 18x3y − 9x2y − 72xy3 + 36y3 Part A: Rewrite the expression so the GCF is factored
vitfil [10]

Answer:

9y(2x - 1)(x - 2y)(x + 2y)

Step-by-step explanation:

Given

18x³y - 9x²y - 72xy³ + 36y³ ← factor out 9y from each term

= 9y(2x³ - x² - 8xy² + 4y²)

Factor the first/second and third/fourth term in the parenthesis

= 9y( x²(2x - 1) - 4y²(2x - 1)) ← factor out (2x - 1) from each term

= 9y(2x - 1)(x² - 4y²)

x² - 4y² is a difference of squares and factors in general as

a² - b² = (a - b)(a + b), thus

x² - 4y²

= x² - (2y)² = (x - 2y)(x + 2y), hence

18x³y - 9x²y + 72xy³ + 36y³

= 9y(2x - 1)(x - 2y)(x + 2y) ← in factored form

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%20%7B%7D%5E%7B2%7D%20-%204%20%7D%7B%20%5Csqrt%7Bx%20%7B%7D%5E%7B2%7D%20-%206x%2
Nimfa-mama [501]

First of all, we can observe that

x^2-6x+9 = (x-3)^2

So the expression becomes

\dfrac{x^2-4}{\sqrt{(x-3)^2}} = \dfrac{x^2-4}{|x-3|}

This means that the expression is defined for every x\neq 3

Now, since the denominator is always positive (when it exists), the fraction can only be positive if the denominator is also positive: we must ask

x^2-4 \geq 0 \iff x\leq -2 \lor x\geq 2

Since we can't accept 3 as an answer, the actual solution set is

(-\infty,-2] \cup [2,3) \cup (3,\infty)

7 0
2 years ago
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