Hello,
Assume Ax+By+Cz+D=0 the plane 's equation.
p=(0,0,0)==>A*0+B*0+C*0+D=0==>D=0
q=(0,1,0)==>A*0+B*1+C*0+0=0==>B=0
r=(1,2,3)==>A*1+0*2+C*3+0=0==>A=-3C
Let C=1==>A=-3
An equation of the plane is -3x+z=0
Answer:

Step-by-step explanation:
We have been given a function
. We are asked to find the zeros of our given function.
To find the zeros of our given function, we will equate our given function by 0 as shown below:

Now, we will factor our equation. We can see that all terms of our equation a common factor that is
.
Upon factoring out
, we will get:

Now, we will split the middle term of our equation into parts, whose sum is
and whose product is
. We know such two numbers are
.




Now, we will use zero product property to find the zeros of our given function.




Therefore, the zeros of our given function are
.
Answer:
7
Step-by-step explanation:
Answer:
6
Step-by-step explanation:
=4!/(4-2)!2!
=4x3x2x1/2x2
=12/2
=6
Answer:
5
Step-by-step explanation:
reminder of rules for division
• If signs are the same then positive result
• If signs are different the negative result
+ 7 ( signs of division are different )
= - 2 + 7
= 5