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Dafna11 [192]
3 years ago
13

How many solutions are there in the given equation? y = 2x² + 6x + 4

Mathematics
1 answer:
tigry1 [53]3 years ago
8 0

Step-by-step explanation:

Use the quadratic formula

=

−

±

2

−

4

√

2

x=\frac{-{\color{#e8710a}{b}} \pm \sqrt{{\color{#e8710a}{b}}^{2}-4{\color{#c92786}{a}}{\color{#129eaf}{c}}}}{2{\color{#c92786}{a}}}

x=2a−b±b2−4ac

Once in standard form, identify a, b and c from the original equation and plug them into the quadratic formula.

2

2

+

6

+

4

=

0

2x^{2}+6x+4=0

2x2+6x+4=0

=

2

a={\color{#c92786}{2}}

a=2

=

6

b={\color{#e8710a}{6}}

b=6

=

4

c={\color{#129eaf}{4}}

c=4

=

−

6

±

6

2

−

4

⋅

2

⋅

4

√

2

⋅

2

x=\frac{-{\color{#e8710a}{6}} \pm \sqrt{{\color{#e8710a}{6}}^{2}-4 \cdot {\color{#c92786}{2}} \cdot {\color{#129eaf}{4}}}}{2 \cdot {\color{#c92786}{2}}}

x=2⋅2−6±62−4⋅2⋅4

brainliest and follow and thanks

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On Monday Sarah had homework in 7∕10 of her classes, Tuesday 3∕5 , Wednesday 9∕11 , and Thursday 1∕2 . Which day did she have th
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It is Wednesday as it has the highest fraction of homework i.e
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Hope this helps :).

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Solve 73 make sure to also define the limits in the parts a and b
Aleks04 [339]

73.

f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}

a)

\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3

b)

Since we can't divide by zero, we need to find when:

x^4-2x^2+144=0

But before, we can factor the numerator and the denominator:

\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}

Now, we can conclude that the vertical asymptotes are located at:

\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}

so, for x = -3:

\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty

For x = 4:

\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty

4 0
1 year ago
Find the center and radius of the circle. Show all work
Alona [7]

The center is (-1, -2) and the radius of the circle is 5.

<h3>What is the equation of a circle?</h3>

The circle equation formula refers to the equation of a circle which represents the center-radius form of the circle.

A circle can be represented as;

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Where; h and k are the center of the circle r is the radius of the circle.

The given equation of the circle is;

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Comparing both the equations;

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Hence, the center is (-1, -2) and the radius of the circle is 5.

Learn more about circle here;

brainly.com/question/11833983

#SPJ2

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Answer:

see explanation

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- 2.3 + (- 5.7)

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The solution is

- 2.3 - 5.7 = - 8

4 0
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