Since the problem is to prove that the two triangles are congruent by applying SSS (side -side -side) congruence theorem, the missing or the additional information that can be shown in the solution is the third side of both triangles must be also equal and congruent. Since in SSS theorem, all sides of a given triangle must be congruent to the opposite three sides of the second triangle.
<span>
<span>
Well,
3/3 * 4/4
1 * 1
So to make a multiplication array of that it would just be one block.
</span>
</span>
<span>
</span>
1. To solve this exercise, you must make a system of equations.
2. You have that f<span>our times a number minus twice another number is -8:
</span>
4x-2y=-8
3. And t<span>he sum of the two number is 19:
</span>
x+y=19
4. As you can see, you have two equations:
4x-2y=-8 (i)
x+y=19 (ii)
5. Let's clear the "x" from the equation (ii):
x=19-y
6. Now, you need to susbtitute x=19-y into the equation (i):
4x-2y=-8
4(19-y)-2y=-8
76-4y-2y=-8
76-6y=-8
-6y=-8-76
-6y=-84
y=-84/-6
y=14
7. You must susbstitute y=14 into the equation (ii) and clear "x":
x+y=19
x+14=19
x=19-14
x=5
The answer is: 5 and 14
Let's simplify step-by-step.
<span><span>12<span>x^2</span></span>+<span>9<span>x^2
</span></span></span>Combine Like Terms:
<span>=<span><span>12<span>x^2</span></span>+<span>9<span>x^2
</span></span></span></span><span>=<span>(<span><span>12<span>x^2</span></span>+<span>9<span>x^2</span></span></span>)
</span></span><span>=<span>21<span>x^<span>2</span></span></span></span>
Given: ax-by>c
Subtract ax on both sides: -by>-ax+c
Divide both sides by -b: y<(-ax+c)/-b
Simplify: y<(ax-c)/b
your answer: y<(ax-c)/b
