I'm assuming you want an equation with a slope of -5 that also passes through the point (1, -7).
An equation that fits this is

Hope this helps!
<u>Prove that:</u>

<u>Proof: </u>
We know that, by Law of Cosines,
<u>Taking</u><u> </u><u>LHS</u>
<em>Substituting</em> the value of <em>cos A, cos B and cos C,</em>



<em>On combining the fractions,</em>

<em>Regrouping the terms,</em>



LHS = RHS proved.
Answer:
No, he cannot
Step-by-step explanation:
Lets say the student takes the maximum socre on his last test, so he gets 100. Now, his average on the six texts will be:
(75+85+90+95+94+100)/6 = 89.83
So, if taking 100 is not enough to get to the average of 90, this means he would need a score greater than 100, which is impossible as 100 is the maximum. So, the student CAN'T get an A
Two pints because there are sixteen ounces in a pint and when you divide 32 by 16 you get 2.
Your anwser to this question is the red its right and i need some points to ask questions!!