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Travka [436]
3 years ago
6

Find three inequalities that are equivalent to 6x + 2x > 6 (x + 2)

Mathematics
1 answer:
lana66690 [7]3 years ago
4 0

Answer:

x > 6

Step-by-step explanation:

6x + 2x > 6 (x + 2)

       8x > 6 (x + 2)

       8x > 6x + 12

      <u>-6x  -6x</u>

       2x/2 > 12/2

             x > 6

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Find the value of angle x , angle y, and angle z in figure.​
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\bf \underline{★ Solution-} \\

In the given figure, we are given with two lines which are parallel to each other. We are also given with two lines which forms a triangle and also forms as a transversal lines to the parallel lines. We are also given that the given triangle is an isosceles triangle. So, we can say that the other angle in the triangle also measures 75°.

Now, let's find the value of the ∠x.

We know that the alternate angles in the parallel line always measures the same as the one which is in it's alternate side. So,

\sf \leadsto \angle{x} = {75}^{\circ}

Now, let's find the value of the ∠z.

We know that, all the angles in a triangle always adds up to 180°. In the given triangle, we are given with two angles, so we can easily find the third angle.

\sf \leadsto {75}^{\circ} + {75}^{\circ} + \angle{z} = {180}^{\circ}

\sf \leadsto {150}^{\circ} + \angle{z} = {180}^{\circ}

\sf \leadsto \angle{z} = 180 - 150

\sf \leadsto \angle{z} = {30}^{\circ}

Now, let's find the value of the ∠y.

We know that all the angles that forms a straight line always equals up to 180° (or) the the straight line angle always measures 180°. So, we can find the value of the ∠y by this concept.

\sf \leadsto {75}^{\circ} + {30}^{\circ} + \angle{y} = {180}^{\circ}

\sf \leadsto {105}^{\circ} + \angle{y} = {180}^{\circ}

\sf \dashrightarrow \angle{y} = 180 - 105

\sf \dashrightarrow \angle{y} = {75}^{\circ}

Therefore,

  1. The value of the ∠x is 75°.
  2. The value of the ∠y is 75°.
  3. The value of the ∠z is 30°.

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Answer:

{x}^{4}  = 2880

Step-by-step explanation:

{y}^{2}  = 20 \: (eq . \: 1)

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