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Vsevolod [243]
3 years ago
6

PLEASE ANSWER THIS 30 POINTS BRAINLIEST WILL BE GIVEN PICTURE INCLUDED!!

Mathematics
2 answers:
Rus_ich [418]3 years ago
8 0

Answer:

a^2 - 9a + 14 = 0 Matches [ 7, 2]

a^2 + 9a + 14 = 0 Matches [ -7,-2]

a^2 + 5a - 14 = 0 Matches [ -7, 2]

a^2  - 5a - 14 = 0 Matches [ 7, -2]

Step-by-step explanation:

I used the diamond method to factor the Equations and the  Solved for x using the factors.

den301095 [7]3 years ago
5 0
The first step is to quickly factor each of the five equations... to do so, find the right factors of the 3rd given number so that they add up in an equal number to the second number... 14 = -7 • -2 and -9 = -7 + -2


a^2 - 9a + 14 = 0
(a - 7) (a - 2)

a - 7 = 0, a = 7
a - 2 = 0, a = 2

{2,7}



a^2 + 9a + 14 = 0
(a + 7) (a + 2)

a + 7 = 0, a = -7
a + 2 = 0, a = -2

{-2, -7}


a^2 + 3a - 10 = 0
(a + 5) (a - 2)

a + 5 = 0, a = -5
a - 2 = 0, a = 2

{-5, 2}


a^2 - 5a - 14 = 0
(a - 7) (a + 2)
a - 7 = 0, a = 7
a + 2 = 0, a = -2

{-2, 7}
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Calculate the limit values:
Nataliya [291]
A) This particular limit is of the indeterminate form,
\frac{ \infty }{ \infty }
if we plug in infinity directly, though it is not a number just to check.

If a limit is in this form, we apply L'Hopital's Rule.

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Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_ {x \rightarrow \infty } \frac{( ln(x ^{2} + 1 ) ) '}{x ' }
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Lim_ {x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ \frac{2x}{x^{2} + 1} }{1}

Still it is of the same indeterminate form, so we apply the rule again,

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b) This limit is also of the indeterminate form,

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we still apply the L'Hopital's Rule,

Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ (tanx)'}{x ' }

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When we plug in zero now we obtain,

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Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ ({e}^{2x} - 1 - 2x)'}{( {x}^{2} ) ' }

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ (2{e}^{2x} - 2)}{ 2x }

It is still of that indeterminate form so we apply the rule again, to obtain;

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ (4{e}^{2x} )}{ 2 }

Now we have remove the discontinuity, we can evaluate the limit now, plugging in zero to obtain;

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = \frac{ (4{e}^{2(0)} )}{ 2 }

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For this kind of question we need to rationalize the radical function, to obtain;

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We now divide both the numerator and denominator by x, to obtain,

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5 0
3 years ago
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6 0
3 years ago
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monitta

Answer:

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Step-by-step explanation:

we require 2 equations with the repeating digits placed after the decimal point.

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7 0
1 year ago
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