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Rom4ik [11]
3 years ago
15

When a quadratic equation has no linear term (variable raised to the first power), the solution becomes simpler. What are the tw

o possible solutions to 5x^2−15=0? Enter your two answers separated by a comma.
Mathematics
1 answer:
OLga [1]3 years ago
3 0

Answer: -\sqrt{3},\ +\sqrt{3}

Step-by-step explanation:

The given quadratic equation :_ 5x^2-15=0

To solve this , we first add 15 on the both the sides of the above equation, we get

5x^2-15+15=0+15\\\\\Rightarrow\ 5x^2=15

Now, we divide both sides by 5, we get

x^2=3

Taking square roots on both the sides , we get

x=\pm\sqrt{3}

Hence, the two possible solutions = -\sqrt{3},\ +\sqrt{3}

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B.

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Select the correct answer. if you roll a single six-sided die, what is the probability of rolling an odd number? a. b. c. d. 3
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\frac{1}{2}

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2 years ago
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ser-zykov [4K]

Answer:

11. 3^2 • 3^5 < 3^8

12. 3^3 • 3^3 > 3^5

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Step-by-step explanation:

11. Which of the following expressions is true?

A.  4^3• 4^4 = 412

4^3• 4^4 = 4^7 = 16384  ❌

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C. 3^2 • 3^5 < 3^8

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12. Which of the following expressions is true?

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4 0
3 years ago
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Sedaia [141]
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x^2+y^2-8x-6y+24=0\quad|-24\\\\(x^2-8x)+(y^2-6y)=-24

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So the answer is C (the same left side of equation).
6 0
3 years ago
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