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Which point is a solution of x^2 + y^2 > 49 and y ≤ –x^2 – 4 ?
A. (–8, –1)
B. (–3, 1)
C. (1, –9)
D. (0, –6)

Which Point Is A Solution Of X2 Y2 Gt 49 And Y X2 4 A 8 1 B 3 1 C 1 9 D 0 6 class=

Sagot :

Ferraitidn

Answer:

C. 1, -9

Step-by-step explanation:

I just put it on desmos

(1, –9) is a solution of [tex]x^{2} +y^{2} > 49[/tex], [tex]y[/tex] ≤ [tex]-x^{2} -4[/tex].

What is inequality?

In mathematics, an inequality is a relation which makes a non-equal comparison between two numbers or other mathematical expressions. It is used most often to compare two numbers on the number line by their size.

Given inequalities

[tex]x^{2} +y^{2} > 49[/tex]..............(1)

[tex]y[/tex] ≤ [tex]-x^{2} -4[/tex]...............(2)

Option (A)

(–8, –1)

Substitute x = -8, y = -1 in equation 1

[tex]x^{2} +y^{2} > 49[/tex]

⇒ [tex](-8)^{2}+(-1)^{2} > 49[/tex]

L.H.S = 65

R.H.S = 49

65 > 49

It is true

Substitute x = -8, y = -1 in equation 2

[tex]y[/tex] ≤  [tex]-x^{2} -4[/tex]

-1 ≤ [tex]-(-8)^{2} -4[/tex]

L.H.S = -1

R.H.S = -68

-1 ≤ -68

It is false.

It is not suitable solution.

Option (B)

(–3, 1)

Substitute x = -3, y = 1 in equation 1

[tex]x^{2} +y^{2} > 49[/tex]

⇒ [tex](-3)^{2}+(1)^{2} > 49[/tex]

L.H.S = 10

R.H.S = 49

10 > 49

It is false

Substitute x = -3, y = 1 in equation 2

[tex]y[/tex] ≤  [tex]-x^{2} -4[/tex]

1 ≤ [tex]-(-3)^{2} -4[/tex]

L.H.S = 1

R.H.S = -13

1 ≤ -13

It is false.

It is not suitable solution.

Option (C)

(1, –9)

Substitute x = 1, y = -9 in equation 1

[tex]x^{2} +y^{2} > 49[/tex]

⇒ [tex]1^{2}+(-9)^{2} > 49[/tex]

L.H.S = 82

R.H.S = 49

82 > 49

It is true

Substitute x = 1, y = -9 in equation 2

[tex]y[/tex] ≤  [tex]-x^{2} -4[/tex]

-9 ≤ [tex](-1)^{2} -4[/tex]

L.H.S = -9

R.H.S = -5

-9 ≤ -5

It is true.

Hence,  (1, –9) is a solution of [tex]x^{2} +y^{2} > 49[/tex], [tex]y[/tex] ≤ [tex]-x^{2} -4[/tex].

Option (D)

(0, –6)

Substitute x = 0, y = -6 in equation 1

[tex]x^{2} +y^{2} > 49[/tex]

⇒ [tex]0^{2}+(-6)^{2} > 49[/tex]

L.H.S = 36

R.H.S = 49

36 > 49

It is false

Substitute x = 0, y = -6 in equation 2

[tex]y[/tex] ≤  [tex]-x^{2} -4[/tex]

-6 ≤ [tex]-(0)^{2} -4[/tex]

L.H.S = -4

R.H.S = -6

-4 ≤ -6

It is false.

It is not suitable solution.

Hence,  (1, –9) is a solution of [tex]x^{2} +y^{2} > 49[/tex], [tex]y[/tex] ≤ [tex]-x^{2} -4[/tex].

Option C is correct.

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