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2 years ago

— 3х + 7 < 19 ? Help plz

Mathematics

2 answers:

Liono4ka [1.6K]2 years ago

8 0

Just like any of your two-step equations, in this inequality,

start by isolating the x term which in this case is -3x by

subtracting 7 from both sides.

This gives us -3x < 12.

Solving from here, we divide both sides by -3.

However, when solving inequalities, you need to watch out.

When you divide both sides of an inequality by a negative number, you must switch the direction of the inequality sign.

Please give this idea your full attention. Even the most advanced algebra students will sometimes forget to switch the direction of the inequality sign when dividing both sides of an inequality by a negative.

So we end up with <em>x > -4</em>.

Maksim231197 [3]2 years ago

6 0

Answer:

x > -4

Step-by-step explanation:

— 3х + 7 < 19

Subtract 7 from each side

— 3х + 7-7 < 19-7

-3x < 12

Divide each side by -3, remembering to flip the inequality

-3x/-3 > 12/-3

x > -4

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Find the perimeter of the quadrilateral that has vertices at (8,13), (5,-18), (-6,-6), and (-2,-5).

kodGreya [7K]

Answer:

72.14 units

Step-by-step explanation:

Let A(8,13), B(5,-18),C(-6,-6) and D(-2,-5)

Apply the distance formula as;

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Finding the lengths of the segments

For AB

$AB=\sqrt{(5-8)^2+(-18-13)^2} =\sqrt{(-3^2)+(-31^2)} =\sqrt{970} =31.14$

For BC

$BC=\sqrt{(-6-5)^2+(-6--18)^2} =\sqrt{(-11^2)+12^2} =\sqrt{265} =16.28$

For CD

$CD=\sqrt{(-2--6)^2+(-5--6)^2} =\sqrt{4^2+1^2} =\sqrt{16+1} =\sqrt{17} =4.12$

For DA

$DA=\sqrt{(8--2)^2+(13--5)^2} =\sqrt{10^2+18^2} =\sqrt{424} =20.6$

Finding the perimeter will be;

31.14+16.28+4.12+20.6=72.14 units

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