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zavuch27 [327]
3 years ago
9

The point (-2, -1) satisfies which of the following inequalities? 5x - 2y + 1 > 0

Mathematics
2 answers:
DanielleElmas [232]3 years ago
6 0

We can plug the given point in each equation and check if it satisfies it

1.

5x - 2y + 1 > 0

5(-2)-2(-1)+1>0

-10+2+1>0

-7>0 (false)

2.

-5x + 2y + 1 > 0

-5(-2) + 2(-1) + 1 > 0

10-2+1>0

9>0

so -5x + 2y + 1 > 0 satisfies the given point (-2,-1)

Ganezh [65]3 years ago
5 0

we are given point (-2, -1)

we can plug this point into inequality and check whether it holds true

(a)

5x - 2y + 1 > 0

now, we can plug x=-2 and y=-1

5*-2 - 2*-1 + 1 > 0

-10 +2 + 1 > 0

-7 > 0

this is not possible

point (-2,-1) does not satisfy inequality

(B)

-5x +2y + 1 > 0

now, we can plug x=-2 and y=-1

-5*-2 +2*-1 + 1 > 0

10 -2 + 1 > 0

9 > 0

this is true

point (-2,-1) does satisfy inequality

(C)

-2x +5y - 1 > 0

now, we can plug x=-2 and y=-1

-2*-2 +5*-1 - 1 > 0

4 -5 - 1 > 0

-2 > 0

this is not true

point (-2,-1) does not satisfy inequality

(D)

2x +5y - 1 > 0

now, we can plug x=-2 and y=-1

2*-2 +5*-1 - 1 > 0

-4 -5 - 1 > 0

-10 > 0

this is not true

point (-2,-1) does not satisfy inequality

so,

option-B.........Answer

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Identify the vertex, axis of symmetry, minimum or maximum, domain, and range of the function ()=−(+)^−

<em><u>Answer:</u></em>

vertex = (-4, -5)

Axis of symmetry = -4

use the (-4, -5) to find the minimum value

Domain = ( - \infty, \infty ) , [ x | x\ is\ real ]\\\\Range = [ -5, \infty ), y\geq -5

<em><u>Solution:</u></em>

Given function is:

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The equation in vertex form is given as:

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Where, (h, k) is constant

On comparing give function with vertex form,

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k = -5

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Axis of symmetry : x co-ordinate of vertex

Thus, axis of symmetry = -4

The coefficient of x^2 is positive in given function.

Thus the vertex point will be a minimum

Minimum\ value = f(\frac{-b}{a})

f(x) = x^2 + 8x + 16 - 5\\\\f(x) = x^2 + 8x + 11

f(x) = ax^2+bx+c

On comparing,

a = 1

b = 8

x = \frac{-b}{2a} = \frac{-8}{2 \times 1} = -4

f(-4) = (-4)^2 + 8(-4) + 11 = 16 - 32 + 11 = -5

Thus, use the (-4, -5) to find the minimum value

Domain and range

f(x) = (x+4)^2 - 5

The domain is the input values shown on the x-axis

The range is the set of possible output values f(x)

Therefore,

Domain = ( - \infty, \infty ) , [ x | x\ is\ real ]\\\\Range = [ -5, \infty ), y\geq -5

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