Hello!
To solve this problem, we will use a system of equations. We will have one number be x and the other y. We will use substitutions to solve for each variable.
x+y=9
x=2y-9
To solve for the two numbers, we need to solve the top equation. The second equation shows that x=2y-9. In the first equation, we can replace 2y-9 for x and solve.
2y-9+y=9
3y-9=9
3y=18
y=6
We now know the value of y. Now we need to find x. We can plug in 6 for y in the second equation to find x.
x=2·6-9
x=12-9
x=3
Just to check, we will plug these two numbers into the first equation.
3+6=9
9=9
Our two numbers are three and six.
I hope this helps!

the equation is a quadratic one, and it has a positive coefficient on the leading term, meaning, is opening upwards, so it has a "burrow" for the vertex.
the minimum or lowest point for a quadratic opening upwards is, well, the vertex point :), the "x" value is the year, the "y" or f(x) value is the population, we're asked for the year, or the x-coordinate of the vertex
well
Answer:
Step-by-step explanation:
From the given information; Let's assume that R should represent the set of all possible outcomes generated from a bit string of length 10 .
So; as each place is fitted with either 0 or 1

Similarly; the event E signifies the randomly generated bit string of length 10 does not contain a 0
Now;
if a 0 bit and a 1 bit are equally likely
The probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 bit and a 1 bit are equally likely is;

so ; if bits string should not contain a 0 and all other places should be occupied by 1; Then:
; 

To get the answer you would:
Multiply 525 by 1.2 525*1.2=
which equals 630
then you would add 525 to 630 525+630=
So the answer would be:
1155
Hope this helps
Answer:
p=1/2*3.14+d
1/2*3.14*19.5=30.615
=30.615+39
=69.615
=69.6
Step-by-step explanation:
since it's half a circle we take the formula of acricle to be half