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Charra [1.4K]
1 year ago
13

20 g of water. specific heat of water is 4.18 j/g°c. temperature changes from 25° c to 20° c, how much heat energy (q) moves fro

m the water to the surroundings?
SAT
1 answer:
Svetach [21]1 year ago
3 0

Answer: I think It's 418 J

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Factor the expression into an equivalent form 12y^2-75.
elena-14-01-66 [18.8K]

Hello oddworld7836!

\huge \boxed{\mathbb{QUESTION} \downarrow}

Factor the expression into an equivalent form 12y² - 75.

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

12 y ^ { 2 } - 75

By observing the expression, we can see that, 3 is the only common factor in both the terms of the expression. So, take the common factor 3 out.

12 y ^ { 2 } - 75 \\  = 3\left(4y^{2}-25\right)

Now, look at (4y² - 25). They don't have any common factors but they appear in the form of the algebraic identity ⇨ a² - b² = (a + b) (a - b). Here,

  • a² = 4, a = 2 (√a² = ✓4 = 2)
  • b² = 25, b = 5 (√b² = ✓25 = 5)

So, the (4y² + 25) becomes...

(4 {y}^{2}  - 25) \\  = \left(2y-5\right)\left(2y+5\right)

Now, bring the 3 (common factor) & rewrite the complete expression.

12 y ^ { 2 } - 75 \\  =  \boxed{ \boxed{ \bf \: 3\left(2y-5\right)\left(2y+5\right) }}

We can't further simplify it. Also, remember that the simplified form of an expression is equivalent to the expression. So, 3 (2y - 5) (2y + 5) is equivalent to 12y² - 75.

__________________

Hope it'll help you!

ℓu¢αzz ッ

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