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2 years ago

Which ordered pair makes both inequalities true?

Mathematics

1 answer:

Tatiana [17]2 years ago

3 0

The ordered pair which makes both inequalities true is: D. (3, 0).

<h3>How to determine ordered pair?</h3>

In Mathematics, an inequality can be used to show the relationship between two (2) or more integers and variables in an equation.

In order to determine ordered pair which makes both inequalities true, we would substitute the points into the inequalities as follows:

At (0, 0), we have:

y > -2x + 3

0 > -2(0) + 3

0 > 3 (false).

y < x – 2

0 < 0 - 2

0 < -2 (false)

At (0, -1), we have:

y > -2x + 3

-1 > -2(0) + 3

-1 > 3 (false).

y < x – 2

-1 < 0 - 2

-1 < -2 (false)

At (1, 1), we have:

y > -2x + 3

1 > -2(1) + 3

1 > -1 (true).

y < x – 2

1 < 1 - 2

1 < -1 (false)

At (3, 0), we have:

y > -2x + 3

0 > -2(3) + 3

0 > -3 (true).

y < x – 2

0 < 3 - 2

0 < 1 (true).

Read more on inequalities here: brainly.com/question/24372553

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The following are the solution to the given points:

Step-by-step explanation:

for point A:

$\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)$

The set A is not part of the subspace $R^3$

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$\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0$

$\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B$

The set B is part of the subspace $R^3$

for point C: $\to C={(x,y,z)|x$

In this, the scalar multiplication can't behold

$\to for (-2,-1,2) \varepsilon C$

$\to -1(-2,-1,2)= (2,1,-1)$ ∉ C

this inequality is not hold

The set C is not a part of the subspace $R^3$

for point D:

$\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)$

The scalar multiplication s is not to hold

$\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)$ ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace $R^3$

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