<span><span><span><span><span>(<span><span>3x</span>−4</span>)</span><span>(<span>2x</span>)</span></span><span>(2)</span></span>+<span>2x</span></span>−1
</span>Distribute:
<span>=<span><span><span><span><span><span>12<span>x2</span></span>+</span>−<span>16x</span></span>+<span>2x</span></span>+</span>−1
</span></span>Combine Like Terms:
<span>=<span><span><span><span>12<span>x2</span></span>+<span>−<span>16x</span></span></span>+<span>2x</span></span>+<span>−1</span></span></span><span>=<span><span><span>(<span>12<span>x2</span></span>)</span>+<span>(<span><span>−<span>16x</span></span>+<span>2x</span></span>)</span></span>+<span>(<span>−1</span>)</span></span></span><span>=<span><span><span>12<span>x2</span></span>+<span>−<span>14x</span></span></span>+<span>−1
</span></span></span>Answer:<span>=<span><span><span>12<span>x2</span></span>−<span>14x</span></span>−<span>1
hope this helps! was there meant to be a parenthesis by the 2 + 2x - 1?? </span></span></span>
Answer:
ASA
Step-by-step explanation:
Given:
Two triangles ABC and EDC such that:
AB ⊥ BD and BD ⊥ DE
C is the midpoint of BD.
The two triangles are drawn below.
Since, AB ⊥ BD and BD ⊥ DE
Therefore, the two triangles are right angled triangle. The triangle ABC is right angled at vertex B. The triangle EDC is right angled at vertex D.
Since, point C is the midpoint of the line segment BD.
Therefore, C divides the line segment BD into two equal parts.
So, segment BC ≅ segment CD (Midpoint theorem)
Now, consider the triangles ABC and EDC.
Statements Reason
1. ∠ABC ≅ ∠CDE Right angles are congruent to each other
2. BC ≅ CD Midpoint theorem. C is midpoint of BD
3. ∠ACB ≅ ∠ECD Vertically opposite angles are congruent
Therefore, the two triangles are congruent by ASA postulate.
So, the second option is correct.
I'm not sure .... but I think you can use cross multiplication to find out the answer.... and then there's another step to finding the compatible numbera
There are 4 hundreds in the number 483
