$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{4}~,~\stackrel{y_1}{3})\qquad K(\stackrel{x_2}{2}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{2+4}{2}~~,~~\cfrac{-3+3}{2} \right)\implies \left( \cfrac{6}{2}~~,~~\cfrac{0}{2} \right)\implies (3~~,~~0)$

6 0

3 years ago

If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum equals 350, which of the following could be equal to n?A.

Svetlanka [38]

Answer:

C. 40

Step-by-step explanation:

Let us say there are x 7's and y 77's.

We have been given that each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum equals 350. We are asked to find the value of n.

Since the sum of all numbers is 350, so we can represent this information in an equation as:

$7x+77y=350$

The only integer solutions for the equation $7x+77y=350$ are 10, 20, 30, 40, 50 .

So, there can be these many terms: 10 or 20 or 30 or 40 or 50 .

The given option only contains 40, so 40 would be the answer. For 40 terms to be there, there has to be 39 sevens and 1 seventy-seven.

Let us verify our answer.

$7x+77y=350$

$7(39)+77(1)=350$

$273+77=350$

$350=350$

Therefore, the total terms (n) could be equal to 40 and option C is the correct choice.

8 0

3 years ago