Answer:
(0,3) is the solution set for given system.
Step-by-step explanation:
Given that:
3x+4y=12 ------- eq1
2x-3y= -9 ------- eq2
Multiplying eq1 with 2 and eq2 with 3 we get:
6x + 8y = 24 --------- eq1
6x - 9y = -27 -------- eq2
Subtracting eq2 from 1 we get:
17y = 51
Dividing both sides by 17 we get:
y = 3
Putting value of 3 in eq1 we get:
3x + 4(3) = 12
3x + 12 = 12
3x = 0
Dividing both sides by 3
x = 0
So the solution set will be: (0,3)
i hope it will help you.
Factor each
60x^4=2*2*3*5*x*x*x*x
45x^5y^5=3*3*5*x*x*x*x*x*y*y*y*y*y
75x^3y=3*5*5*x*x*x*y
common is 3*5*x*x*x=15x^3
gcf=15x^3
From a formula located here:
http://www.1728.org/quadltrl.htm
we see that
<span>4 • Side² = Long Diagonal² + Short Diagonal²
Long Diagonal = 24
Short Diagonal = 10
</span><span>4 • Side² = 24^2 + 10^2 </span>
<span>4 • Side² = 576 + 100
</span><span>4 • Side² = 676
</span><span><span>Side²</span> = 169
Side (or line AB) = 13
</span>