Compute $-8\cdot 4-(-6\cdot -3)+(-10\cdot -5)$.
The order of operations says that we must perform the multiplication before the addition and subtraction.  Recall that ``negative times positive equals negative" and ``negative times negative equals positive".  We obtain \begin{align*}
-8\cdot 4-(-6\cdot -3)+(-10\cdot -5)&=-32-18+50\\
&=-(32+18)+50\\
&=-50+50 \\
&=50+(-50) \\
&=50-50 \\
&=\boxed{0}.
\end{align*}
0
Level 2
Prealgebra
Problem: Compute $-8\cdot 4-(-6\cdot -3)+(-10\cdot -5)$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the expression, first evaluate each multiplication term separately. 

Calculate $-8 \cdot 4$: 
- Multiply $-8$ by $4$ to get $-32$.

Next, calculate $-6 \cdot -3$: 
- Multiply $-6$ by $-3$. The product of two negative numbers is positive, so $-6 \cdot -3 = 18$.

Then, calculate $-10 \cdot -5$: 
- Again, multiply two negative numbers. Thus, $-10 \cdot -5 = 50$.

Now, substitute these results back into the original expression:
- The expression becomes $-32 - 18 + 50$.

Proceed with the addition and subtraction from left to right:
- First, compute $-32 - 18$. This results in $-50$.
- Next, add $50$ to $-50$: $-50 + 50 = 0$.

Thus, the final answer is $0$.<|END_OF_LATENT|> The final answer is $0$.