Answer:
3. One Solution. (0,4). (they have different slopes and y-intercepts)
The solution can be found using elimination method:
Add the equations together
y = -x + 4
y = x - 4
2y = 0
y = 0
Solve for x
0 = x - 4
x = 4
(4, 0)
4. No solution
x + y = -2 and 3 which is not possible
The two lines would be parallel
Answer:
27
Step-by-step explanation:
If a= -3 what is 3a^2 = 27
To solve, simply layout an equation.
(2x+3)(x+4)=1
Step 1: Simplify both sides of the equation.
2x2+11x+12=1
Step 2: Subtract 1 from both sides.
2x2+11x+12−1=1−1
2x2+11x+11=0
Step 3: Use quadratic formula with a=2, b=11, c=11.
x=-b±√b^2-4ac/2a
So, the answer for this problem is:
x=-11/4+1/4√33 or -11/4+-1/4√33
Answer:
<u><em>note:</em></u>
<u><em>solution is attached</em></u>
Given that JKLM is a rhombus and the length of diagonal KM=10 na d JL=24, the perimeter will be found as follows;
the length of one side of the rhombus will be given by Pythagorean theorem, the reason being at the point the diagonals intersect, they form a perpendicular angles;
thus
c^2=a^2+b^2
hence;
c^2=5^2+12^2
c^2=144+25
c^2=169
thus;
c=sqrt169
c=13 units;
thus the perimeter of the rhombus will be:
P=L+L+L+L
P=13+13+13+13
P=52 units
