No two <em>intergers</em> can solve this problem. If you're not in Algebra II yet, the answer is probably "none."
Just look at the factor pairs of 25:
1 × 25
-1 × 25
5 × 5
-5 × -5
Clearly none of those can add up to 6.
If you want the more complex answer, I'll show you how here. If you don't understand why it doesn't work, that's okay. I just want you to see that there's not an actual answer to the problem.
a+b = 6
a = 6-b
ab = 25
(6-b)b = 25
6b -b² = 25
-b² + 6b = 25
b² -6b = -25
Factor by splitting the middle.
Half of -6 is -3, (-3)² = 9. Add this to each side.
b² -6b + 9 = -16
Factor the perfect square trinomial.
(b-3)² = -16
Take the square root of each side.
b-3 = 4i
b = 3+4i
a+b = 6
a+3+4i = 6
a= 3-4i
<em>(The "i" stands for an imaginary number, specifically, the square root of -1.)</em>
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
Answer:
Step-by-step explanation:
(A) Definition of Supplementary angles
(B) Distributive property
(C) Transitive property
(D) Reflexcive property
(E) Division property of equality
Here' s the equation for one year:
0.025(3000) + 3000
Because it is 5 years:
5(0.025(3000)) + 3000
0.075(3000) + 3000
We can make it simpler:
1.075(3000)
Multiply:
3225
You will have $3225