Answer:
A) x = 3 or -1
B) x = -7
C)x = -7
Step-by-step explanation:
A) x² + 2x + 1 = 2x² - 2
Rearranging, we have;
2x² - x² - 2x - 2 - 1 = 0
x² - 2x - 3 = 0
Using quadratic formula, we have;
x = [-(-2) ± √((-2)² - 4(1 × -3))]/(2 × 1)
x = (2 ± √16)/2
x = (2 + 4)/2 or (2 - 4)/2
x = 6/2 or -2/2
x = 3 or -1
B) ((x + 2)/3) - 2/15 = (x - 2)/5
Multiply through by 15 to get;
5(x + 2) - 2 = 3(x - 2)
5x + 10 - 2 = 3x - 6
5x - 3x = -6 - 10 + 2
2x = -14
x = -14/2
x = -7
C) log(2x + 3) = 2log x
From log derivations, 2 log x is same as log x²
Thus;
log(2x + 3) = logx²
Log will cancel out to give;
2x + 3 = x²
x² - 2x - 3 = 0
Using quadratic formula, we have;
x = [-(-2) ± √((-2)² - 4(1 × -3))]/(2 × 1)
x = (2 ± √16)/2
x = (2 + 4)/2 or (2 - 4)/2
x = 6/2 or -2/2
x = 3 or -1
Answer:
4 .For 32 it goes into 4 and for 12 it goes into 4
Step-by-step explanation:
The value of the coin times the number of coins all added up equals the total value of the piggy bank
so
0.05n + 0.10d = 5.60
then they have given us "number of nickels is 8 less than 3 times the dimes"
this translates to
n = 3d - 8
we plug this into our first equation and we get
0.05(3d - 8) + 0.10d = 5.60
now we solve for d
first we can distribute the 0.05
0.15d - 0.40 + 0.10d = 5.60
then we can add the 0.15d and 0.10d together
and get
0.25d - 0.40 = 5.60
then we can add 0.40 to both sides
0.25d = 6.00
then divide both sides by 0.25
d = 6.00 / 0.25 = 24
so we have 24 dimes
then to find nickels, we plug in 24 for d into
n = 3d - 8
and get
n = 3(24) - 8
n = 72 - 8
n = 64
therefore we have
64 nickels
and
24 dimes
9514 1404 393
Answer:
- ABHGEFDCA
- does not exist
- ECBADFE
Step-by-step explanation:
A Hamiltonian circuit visits each node once and returns to its start. There is no simple way to determine if such a circuit exists.
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For graph 2, if there were a circuit, paths ACB, ADB, and AEB would all have to be on it. Inclusion of all of those requires visiting nodes A and B more than once, so the circuit cannot exist.
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For graph 3, the circuit must include paths BAD and DFE. That only leaves node C, which can be reached from both nodes B and E, so path ECB completes the circuit.
