Question

question 7 (1 point) find m∠bdc question 8 (1 point) tell whether ∠1 and ∠2 are only adjacent, adjacent and form a linear pair, or not adjacent.

Explanation:

Step1: Use linear - pair property

Since $\angle BDC$ and $\angle CDA$ form a linear - pair, their sum is $180^{\circ}$. So, $(-3x + 20)+(-2x + 55)=180$.

Step2: Combine like - terms

Combine the $x$ terms and the constant terms: $(-3x-2x)+(20 + 55)=180$, which simplifies to $-5x+75 = 180$.

Step3: Solve for $x$

Subtract 75 from both sides: $-5x=180 - 75$, so $-5x=105$. Then divide both sides by $-5$, we get $x=-21$.

Step4: Find $m\angle BDC$

Substitute $x = - 21$ into the expression for $\angle BDC$: $m\angle BDC=-3x + 20=-3\times(-21)+20=63 + 20=83$. But this is wrong. Let's assume the angle at $D$ is a right - angle, so $(-3x + 20)+(-2x + 55)=90$.

Step5: Combine like - terms for new equation

$(-3x-2x)+(20 + 55)=90$, which simplifies to $-5x+75 = 90$.

Step6: Solve for $x$ in new equation

Subtract 75 from both sides: $-5x=90 - 75$, so $-5x = 15$. Divide both sides by $-5$, we get $x=-3$.

Step7: Find $m\angle BDC$

[LLM SSE On Failure]

Answer:

Step1: Use linear - pair property

Since $\angle BDC$ and $\angle CDA$ form a linear - pair, their sum is $180^{\circ}$. So, $(-3x + 20)+(-2x + 55)=180$.

Step2: Combine like - terms

Combine the $x$ terms and the constant terms: $(-3x-2x)+(20 + 55)=180$, which simplifies to $-5x+75 = 180$.

Step3: Solve for $x$

Subtract 75 from both sides: $-5x=180 - 75$, so $-5x=105$. Then divide both sides by $-5$, we get $x=-21$.

Step4: Find $m\angle BDC$

Substitute $x = - 21$ into the expression for $\angle BDC$: $m\angle BDC=-3x + 20=-3\times(-21)+20=63 + 20=83$. But this is wrong. Let's assume the angle at $D$ is a right - angle, so $(-3x + 20)+(-2x + 55)=90$.

Step5: Combine like - terms for new equation

$(-3x-2x)+(20 + 55)=90$, which simplifies to $-5x+75 = 90$.

Step6: Solve for $x$ in new equation

Subtract 75 from both sides: $-5x=90 - 75$, so $-5x = 15$. Divide both sides by $-5$, we get $x=-3$.

Step7: Find $m\angle BDC$

[LLM SSE On Failure]