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k0ka [10]
3 years ago
6

Fins the zeroes of the quadratic equation : 4√ 5 x^ 2 - 24x - 9√ 5

Mathematics
1 answer:
liubo4ka [24]3 years ago
5 0

Answer:

Step-by-step explanation:

Hello,

First of all, we know that the solution of the following equation

   ax^2+bx+c=0

are

   \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} \ \text{ when } \Delta=b^2-4ac \geq 0

<em>I would suggest that you try to apply this formula first and check the solution only after you try.</em>

Let's apply it in this case, we have:

a=4\sqrt{5} \\ \\ b=-24 \\ \\ c= -9\sqrt{5}

\Delta=b^2-4ac=24^2+4*9*5=1296 \\ \\ \sqrt{\Delta}=36 \ \text{ and the solutions are } \\ \\ \\ x_1=\dfrac{24-36}{8\sqrt{5}}=\dfrac{-12}{8\sqrt{5}}=\boxed{\dfrac{-3}{2\sqrt{5}}} \\ \\x_2=\dfrac{24+36}{8\sqrt{5}}=\dfrac{60}{8\sqrt{5}}=\boxed{\dfrac{15}{2\sqrt{5}}}

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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Answer:

23

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14 + 9 = 23

-14 + 23 = 9

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2 years ago
Help!!!!!!!!!! TYSM if you answer this!!!!!!!!!!!
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10p+5

Step-by-step explanation:

8 0
2 years ago
I have A and B, I need help with C.
WARRIOR [948]
The answer is $382.80. the perimeter is 600ft. if fence post are placed every five feet, you need 120 post because 600÷5 is 120. each post cost 3.19 so you need 382.80 because 3.19×120 is 382.8
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3 years ago
Which of the following is a trinomial?
Sedaia [141]

Answer:

A) c^2 + c + 6

Step-by-step explanation:

We have been given four choices out of which we need to select the choice which is trinomial.

Trinomial means polynomial having three terms.

A) c2 + c + 6

It has 3 terms.

B) c2 − 16

It has 2 terms.

C) −8c

It has 1 term.

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It has 4 terms.

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Hence correct choice is A) c^2 + c + 6


5 0
3 years ago
Find the length of the curve y = 3/5x^5/3 - 3/4x^1/3 + 6 for 1 &lt; = x &lt; = 8. The length of the curve is . (Type an exact an
Mashutka [201]

Answer:

\sqrt\frac{387}{20}

Step-by-step explanation:

Arc Length =\int\limits^a_b {\sqrt{1+(\frac{dy}{dx})^2 } } \, dx

y=\dfrac{3}{5}x^{\frac{5}{3}}-  \dfrac{3}{4}x^{\frac{1}{3}}+6

\frac{dy}{dx} =x^{\frac{2}{3}}-\dfrac{1}{4}x^{-\frac{2}{3}}

1+(\frac{dy}{dx})^2 }=1+(x^{\frac{2}{3}}-\dfrac{1}{4}x^{-\frac{2}{3}})^2\\=1+(x^{\frac{4}{3}}-\dfrac{1}{2}+ \dfrac{1}{16}x^{-\frac{4}{3}})

=\dfrac{1}{2}+x^{\frac{4}{3}}+ \dfrac{1}{16}x^{-\frac{4}{3}}

For the Interval 1\leq x\leq 8

Length of the Curve =\int\limits^8_1 {\sqrt{\dfrac{1}{2}+x^{\frac{4}{3}}+ \dfrac{1}{16}x^{-\frac{4}{3}} } } \, dx\\

Using T1-Calculator

=\sqrt\frac{387}{20}

3 0
3 years ago
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