The missing value is 12 in a system of equations with infinitely many solutions conditions.
It is given that in the system of equations there are two equations given:
It is required to find the missing value in the second equation.
<h3>What is a linear equation?</h3>
It is defined as the relation between two variables if we plot the graph of the linear equation we will get a straight line.
We have equations:
Let's suppose the missing value is 'Z'
We know that the two pairs of equations have infinitely many solutions if and if they have the same coefficients of variables and the same constant on both sides.
From equation (1)
(multiply both the sides by 3)
...(3)
By comparing the equation (2) and (3), we get
M = 12
Thus, the missing value is 12 in a system of equations with infinitely many solutions conditions.
Learn more about the linear equation.
brainly.com/question/11897796
Hello,
A: roots: -1,-3
a point (-2,1)
Vertex=((-2,1)
y=k*(x+1)(x+3) using roots
but k*(-2+1)(-2+3)=1==>k*(-1)*1=1==>k=-1
eq: y=-(x+1)(x+3)
==>y=-(x²+3x+x+3)
==>y=-x²-4x-3
y=k(x+2)²+1 if x=-1,y=0 ==>k*1+1=0==>k=-1
==>y=-(x+2)²+1
Answer :A--> R,K
B)
y=k(x+4)²-2 and k=-1/2
y=-1/2(x+4)²-2
y=-1/2x²-4x-10
answer B--> I,≈W if it is written -1/2*x² (square has been forgotten)
C:
y=2x²-16x+30
y=2(x-4)²-2
answer : C-->S,J
D:
y=-(x+3)(x+1)
y=-x²-4x-3
=-(x+2)²+1
answer D--> V,L
E:
Here there is a problem: or the graph is wrong, or 2 equations are missing!
y=1(x+1)(x-3) using roots
y=x²-2x-3 ≈ T si it were -2x and not +2x.
y=(x-1)²-4 ≈H is it were -1 in place of +1 [H:y=(x+1)²-4]
Answer:
-2 1/5, -1 2/5, 1/5, |-3/5|, |-1 1/5| ,|-2 2/5|
Step-by-step explanation:
Absolute value means take the positive value
-2 1/5
|-1 1/5| = 1 1/5
|-2 2/5| = 2 2/5
1/5
|-3/5| = 3/5
-1 2/5
We want the numbers from smallest to largest
The most negative is
-2 1/5, -1 2/5
Then to the fractions
1/5, |-3/5|
Then the positive numbers
|-1 1/5| ,|-2 2/5|
Consider this option:
Volume is 256π/5, area is 32/3.
Details are in the attachment.
P.S. intersection points are (-2;4) and (2;4).
Start by factoring an x term out of the equation:
x(x-11)=0
Now, you can set each term equal to zero and solve for x:
x=0
x-11=0
x=11
x={0, 11}
Hope this helps!!
