Answer:
I think it's the last one! In an experiment, one group is studied over a short period of time---
Answer:
x=-6 is an extraneous solution
Step-by-step explanation:
we have
squared both sides
solve the quadratic equation by graphing
The solution is x=-6 and x=-1
see the attached figure
<u><em>Verify each solution</em></u>
substitute the value of x in the original expression
For x=-6
----> is not true
so
Is an extraneous solution
For x=-1
----> is true
so
Is the solution
(2x/-5x)+x^2
(x(2)/(x(-5))+x^2
(-2/5)+x^2, the x values cancel in numerator and denominator and simplify to -2/5
Answer:
Yes they will intersect
Function 1= F(X)=2X+5
Function 2=H(X)=3X+2
INTERSECT=(3,11)
Step-by-step explanation:
First of all, we create 2 LINEAR function, i created the function f(x)=2x+5 and the function h(x)=3x+2, both are linear(without a quadratic term). Then
you replace the x for a number:
Table 1 (F(X)=2X+5) Table 2 (H(X)=3X+2)
X=1----->Y=2+5=7 X=1------>Y=3·1+2=5
X=2---->Y=2·2+5=9 X=2----->Y=3·2+2=8
X=3---->Y=3·3+5=11 X=3----->Y=3·3+2=11
With both tables of data we can see that in the X=3/Y=11 point this two linear functions will intersect so the answer is that the two functions will intersect at (3,11)----->(X,Y)
