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Alona [7]
3 years ago
7

Solve the following quadratic equation using the quadratic formula. Which of the following expressions gives the numerators of t

he solutions?
10x^2 - 19x + 6 = 0
Mathematics
2 answers:
Illusion [34]3 years ago
5 0

I hope the choices for the numerators of the solutions are given.

I am showing the complete work to find the solutions of this equation , it will help you to find an answer of your question based on this solution.

The standard form of a quadratic equation is :

ax² + bx + c = 0

And the quadratic formula is:

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

So, first step is to compare the given equation with the above equation to get the value of a, b and c.

So, a = 10, b = -19 and c = 6.

Next step is to plug in these values in the above formula. Therefore,

x=\frac{(-19)-\pm\sqrt{(-19)^2-4*10*6}}{2*10}

=\frac{19\pm\sqrt{361-240}}{20}

=\frac{19\pm\sqrt{121}}{20}

=\frac{19\pm11}{20}

So, x=\frac{19-11}{20} ,\frac{19+11}{20}

x=\frac{8}{20} , \frac{30}{20}

So, x= \frac{2}{5} ,\frac{3}{2}

Hope this helps you!

RideAnS [48]3 years ago
3 0

<em>This is the answer:</em>

19 ± 11

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PLEASE HELP!!!! 20 POINTS!!!!
galben [10]
These are two questions and two answers.

Question 1) Which of the following polar equations is equivalent to the parametric equations below? 

<span> x=t²
y=2t</span>

Answer: option <span>A.) r = 4cot(theta)csc(theta)
</span>

Explanation:

1) Polar coordinates ⇒ x = r cosθ and y = r sinθ

2) replace x and y in the parametric equations:

r cosθ = t²
r sinθ = 2t

3) work r sinθ = 2t

r sinθ/2 = t 
(r sinθ / 2)² = t²

4) equal both expressions for t²

r cos θ = (r sin θ / 2 )²

5) simplify

r cos θ = r² (sin θ)² / 4

4 = r (sinθ)² / cos θ

r = 4 cosθ / (sinθ)²

r = 4 cot θ csc θ ↔ which is the option A.


Question 2) Which polar equation is equivalent to the parametric equations below?

<span> x=sin(theta)cos(theta)+cos(theta)
y=sin^2(theta)+sin(theta)</span>

Answer: option B) r = sinθ + 1


Explanation:

1) Polar coordinates ⇒ x = r cosθ, and y = r sinθ

2) replace x and y in the parametric equations:

a) r cosθ = sin(θ)cos(θ)+cos(θ)
<span> b) r sinθ =sin²(θ)+sin(θ)</span>

3) work both equations

a) r cosθ = sin(θ)cos(θ)+cos(θ) ⇒ r cosθ = cosθ [ sin θ + 1]  ⇒ r = sinθ + 1


<span> b) r sinθ =sin²(θ)+sin(θ) ⇒ r sinθ = sinθ [sinθ + 1] ⇒ r = sinθ + 1
</span><span>
</span><span>
</span>Therefore, the answer is r = sinθ + 1 which is the option B.
6 0
3 years ago
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Citrus2011 [14]
(x+12)(x-3) Try this answer hopefully it is correct.
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2 years ago
Can someone help me with this guys? Thank you so much! I mark as brainliest
Vitek1552 [10]

Answer:

y=45

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Step-by-step explanation:

This is a right isosceles triangle, so

y=45

x=9

z= 9 root 2

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81+81=z²

162=z²

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5 0
3 years ago
Find the value of x in the figure below 13x+8 2x+7
liberstina [14]

Step-by-step explanation:

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step 2. 15x + 15 = 180 (like terms)

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3 years ago
Find the equation, f(x) = a(x-h)2 + k, for a parabola that passes through the point (0, 0) and has (-3, -6) as its vertex. What
Arisa [49]
F(x) = a(x-h)²<span> + k

</span><span><u>Given that the vertex is (-3 -6):</u>

</span>f(x) = a(x + 3)² -6
<span>
<u>Given that it passes through (0.0), find a:</u>

a(0 + 3)</span>² - 6 = 0
<span>
9a - 6 = 0

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a = 6/9 =2/3

<u>So the equation is :</u>

</span>f(x) = 2/3(x + 3)² -6
<span>
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</span>f(x) = 2/3(x² + 6x + 9) - 6

f(x) = 2/3x² + 4x + 6 - 6

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Answer: </span>f(x) = 2/3x² + 4x<span>


</span>
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