Answer:
2/4
Step-by-step explanation:
Answer:
Simple the area is 88
Step-by-step explanation:
Answer:
We can do it with envelopes with amounts $1,$2,$4,$8,$16,$32,$64,$128,$256 and $489
Step-by-step explanation:
- Observe that, in binary system, 1023=1111111111. That is, with 10 digits we can express up to number 1023.
This give us the idea to put in each envelope an amount of money equal to the positional value of each digit in the representation of 1023. That is, we will put the bills in envelopes with amounts of money equal to $1,$2,$4,$8,$16,$32,$64,$128,$256 and $512.
However, a little modification must be done, since we do not have $1023, only $1,000. To solve this, the last envelope should have $489 instead of 512.
Observe that:
- 1+2+4+8+16+32+64+128+256+489=1000
- Since each one of the first 9 envelopes represents a position in a binary system, we can represent every natural number from zero up to 511.
- If we want to give an amount "x" which is greater than $511, we can use our $489 envelope. Then we would just need to combine the other 9 to obtain x-489 dollars. Since
, by 2) we know that this would be possible.
Something funny is that the x value of the vertex lies directl in the middle of the x intercepts
so
we see the x intercepts or 0's at x=8 and 2
the average is x=5
so find f(5) to find the y value of the vertex
f(5)=(5-8)(5-2)
f(5)=(-3)(3)
f(5)=-9
vertex is at (5,-9)
the actual way the teacher wants is to expand then compltete the square to get into the form f(x)=a(x-h)^2+k where the vertex is (h,k)
but whatever
verrtex is at (5,-9)
The answer is C (EF and HI)
This is because both lines pass through the centre A, meaning both are diameters. Hence, they are equal in length.