Answer:
Hopes it helps
Step-by-step explanation:
The Quadratic Polynomial is
2 x² +x -4=0
Using the Determinant method to find the roots of this equation
For, the Quadratic equation , ax²+ b x+c=0
(b) x²+x=0
x × (x+1)=0
x=0 ∧ x+1=0
x=0 ∧ x= -1
You can look the problem in other way
the two Quadratic polynomials are
2 x²+x-4=0, ∧ x²+x=0
x²= -x
So, 2 x²+x-4=0,
→ -2 x+x-4=0
→ -x -4=0
→x= -4
∨
x² +x² +x-4=0
x²+0-4=0→→x²+x=0
→x²=4
x=√4
x=2 ∧ x=-2
As, you will put these values into the equation, you will find that these values does not satisfy both the equations.
So, there is no solution.
You can solve these two equation graphically also.
Hmm...I tried to divide 29 by 14, and I got 2.07142857143.... Does it let you round the decimal..?
Answer:
3448.5
Step-by-step explanation:
I=PRT
I=22990×0.05×3
=$3448.5
Answer:
66.46
Step-by-step explanation:
Applying the rule of modulus for given values, the expression will result to 20.
<h3>What is Modulus or Absolute Values</h3>
The modulus of a value for example |a|=a if a is greater than or equal to zero
and also
|a|=-a if a is less than zero
In the expression given; the modulus of -6 written as
|-6|=-(-6) this is because the value is less than zero
The modulus of 2 written as; |2|=2 this is because the value is greater than zero. and;
|The modulus of -14 written as; |-14|= -(-14) since the value is less than zero.
Hence, we can rewrite the expression as;
we deal with multiplication first following the rules of BODMAS
2×-(-6)-3×2+[-(-14)]
2×6-6+14
and thus adding values
12+14-6
and finally we subtract;
26-6
which will results to the solution of 20.