We know that
<span>the sum of the residuals must be equal to zero
</span>let
x------> <span>the third residual
-12+24+x=0
12+x=0-----> add -12 both sides
x=-12
the answer is
</span> the third residual is -12<span>
</span>
The answer is x=1
.........
Answer:
3156
Step-by-step explanation:
- <em>Used formula:</em>
- <em>(1² + 2² + 3² + ... + n²) =1/6*n(n + 1)(2n + 1)</em>
--------
- 10²+12²+14²+......+26² =
- (2*5)²+(2*6)² + (2*7)² + ... + (2*13)² =
- 4*(5²+6²+7²+...+13²) =
- 4*(1²+2²+...+13² - (1²+2²+3²+4²)) =
- 4*(1/6*13(13+1)(2*13+1) - (1+4+9+16)) =
- 4*(1/6*13*14*27- 30) =
- 4*(819 - 30) =
- 4*789 =
- 3156
Let
x-------> the first number
y-------> the second number
P-----> product
we know that
x+2y=56----------> 2y=56-x-----> y=28-0.5x------> equation 1
P=x*y----> equation 2
substitute equation 1 in equation 2
P=x*[28-0.5x]-----> P=28x-0.5x²
using a graph tool
see the attached figure
the vertex is the point (28,392)
that means
for x=28
the product is 392 (maximum)
find the value of y
y=28-0.5x----> y=28-0.5*28----> y=14
the answer isthe numbers are 28 and 14