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Sav [38]
3 years ago
15

Solve the quadratic equation give your answer to 2 decimal places : 3x^2+x-5=0

Mathematics
1 answer:
Dima020 [189]3 years ago
3 0

Given:

The quadratic equation is:

3x^2+x-5=0

To find:

The solution for the given equation rounded to 2 decimal places.

Solution:

Quadratic formula: If a quadratic equation is ax^2+bx+c=0, then:

x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

We have,

3x^2+x-5=0

Here, a=3,b=1,c=-5. Using the quadratic formula, we get

x=\dfrac{-1\pm \sqrt{1^2-4(3)(-5)}}{2(3)}

x=\dfrac{-1\pm \sqrt{1+60}}{6}

x=\dfrac{-1\pm \sqrt{61}}{6}

x=\dfrac{-1\pm 7.81025}{6}

Now,

x=\dfrac{-1+7.81025}{6}

x=1.13504167

x\approx 1.14

And

x=\dfrac{-1-7.81025}{6}

x=-1.468375

x\approx -1.47

Therefore, the required solutions are 1.14 and -1.47.

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Answer:

\boxed {\tt 69.08 \ centimeters}

Step-by-step explanation:

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13.......................
Alla [95]

Answer:

The answer is (d) ⇒ pq^{2}r\sqrt[3]{pr^{2}}

Step-by-step explanation:

* To simplify the cube roots:

 If its number then the number must be written in the form x³

 then we divide the power by 3 to cancel the radical

 If its variable we divide its power by 3 to cancel the radical

∵ \sqrt[3]{p^{4}q^{6}r^{5}}=p^{\frac{4}{3}}q^{\frac{6}{3}}r^{\frac{5}{3}}}

∴ p^{\frac{4}{3}}q^{2}}r^{\frac{5}{3}}=p^{1\frac{1}{3}}q^{2}r^{1\frac{2}{3}}

∵ p^{\frac{1}{3}}=\sqrt[3]{p}

∵ r^{\frac{2}{3}}=\sqrt[3]{r^{2}}

∴ p(p)^{\frac{1}{3}}q^{2}r(r)^{\frac{2}{3}}=p(\sqrt[3]{p})q^{2}r(\sqrt[3]{r^{2}})

∴ prq^{2}\sqrt[3]{pr^{2}}}

∴ The answer is (d)

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Help help and this is a test btw pls yall ​
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Answer:

Hiiiii

This is the answer

Hope it's help

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