the assumption being that the first machine is the one on the left-hand-side and the second is the one on the right-hand-side.
the input goes to the 1st machine and the output of that goes to the 2nd machine.
a)
if she uses and input of 6 on the 2nd one, the result will be 6² - 6 = 30, if we feed that to the 1st one the result will be √( 30 - 5) = √25 = 5, so, simply having the machines swap places will work to get a final output of 5.
b)
clearly we can never get an output of -5 from a square root, however we can from the quadratic one, the 2nd machine/equation.
let's check something, we need a -5 on the 2nd, so
so if we use a "1" as the output on the first machine, we should be able to find out what input we need, let's do that.
so if we use an input of 6 on the first machine, we should be able to get a -5 as final output from the 2nd machine.
Answer:
AD = 84
Step-by-step explanation:
Since B is the midpoint of AC then AB = BC = 2x - 5
Since C is the midpoint of AD then AC = CD, thus
AB + BC = CD, that is
2x - 5 + 2x - 5 = x + 29
4x - 10 = x + 29 ( subtract x from both sides )
3x - 10 = 29 ( add 10 to both sides )
3x = 39 ( divide both sides by 3 )
x = 13
Hence
AD = AB + BC + CD
= 2x - 5 + 2x - 5 + x + 29 = 5x + 19, thus
AD = (5 × 13) + 19 = 65 + 19 = 84
Answer:
<BCO = <BAO = 20degrees
Step-by-step explanation:
If <ABC measures 100 and is inscribed in a circle O. find <BAO and <BCO
To get <BAO and <BCO, we need to get <AOC first.
From the figure, it can be seen that triangle ABC is an isosceles trinagle. Hence;
<BAC + <BCA + 100 = 180
Since <BAC = <BCA
<BAC + <BAC = 180 - 100
2<BAC = 80
<BAC = 80/2
<BAC = 40
Also;
<BAO = <BCO and <BAO = <BAC/2
<BAO = 40/2 = <BCO
Hence <BCO = <BAO = 20degrees
Since DB bisects ∠ABC and ∠ABC= ∠ABD + ∠DBC, then
∠ABD = ∠DBC, that is
4x = x + 36 ( subtract x from both sides )
3x = 36 ( divide both sides by 3 )
x = 12
∠ABD = 4x = 4× 12 = 48°
∠DBC= x + 36 = 12 + 36 = 48°
∠ABC = 48° + 48° = 96°
Thus statement A is FALSE
statement B is TRUE
statement C is TRUE
