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Furkat [3]
3 years ago
8

What is the perimeter of abc with vertices a(-2 9)?

Mathematics
1 answer:
vazorg [7]3 years ago
6 0
For finding the perimeter of abc (triangle), we MUST need the two others coordinates of b and c
let b(1, 2) and c (2, -4)
we need to calculate the distances ab, ac, and bc
vector (ab)=(1-(-2), 2-9)=(3, -7), so its length is ab=sqrt(3²+ (-7)²)=7.61
realising the same method, we find bc=6.08, ac=13.60

so the perimeter of abc is P= ab+bc+ca=13.60+6.08+7.61=27.29
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Answer:

See the explanation.

Step-by-step explanation:

We are given the function f(x) = x² + 2x - 5

Zeros :

If f(x) = 0 i.e. x² + 2x - 5 = 0

The left hand side can not be factorized. Hence, use Sridhar Acharya formula and  

x= \frac{-2+\sqrt{2^{2}-4\times(-5)\times1 } }{2} and  

x= \frac{-2-\sqrt{2^{2}-4\times(-5)\times1 } }{2}

⇒ x = -3.45 and 1.45

Y- intercept :

Putting x = 0, we get, f(x) = - 5, Hence, y-intercept is -5.

Maximum point :

Not defined

Minimum point:  

The equation can be expressed as (x + 1)² = (y + 5)

This is an equation of parabola having the vertex at (-1,-5) and axis parallel to + y-axis

Therefore, the minimum point is (-1,-5)

Domain :  

x can be any real number

Range:  

f(x) ≥ - 6

Interval of increase:

Since this is a parabola having the vertex at (-1,-5) and axis parallel to + y-axis.

Therefore, interval of increase is +∞ > x > -1

Interval of decrease:

-∞ < x < -1

End behavior :  

f(x) = x^{2} +2x-5 =x^{2}  (1+\frac{2}{x} -\frac{5}{x^{2} } )

So, as x tends to +∞ , then f(x) tends to +∞

And as x tends to -∞, then f(x) tends to +∞. (Answer)

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3 years ago
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Step-by-step explanation:

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