No nothing is even it is supposed to have 2 pairs of congruent and parallel sides
Answer:
c, e, f, and i
Step-by-step explanation:
a. is bounded above at y = 1 and below at y = 0
b. is unbounded both above and below
c. is bounded below at y = 0 and unbounded above
d. is unbounded both above and below
e. is bounded below at y = 0 and unbounded above
f. is bounded below at y = 0 and unbounded above
g. is bounded below at y = 0 and bounded above at y = 1
h. is unbounded both above and below
i. is bounded below at y = 0 and unbounded above
j. is unbounded both above and below
Answer:
* Elimination; a coefficient in Equation I is an integer multiple of a coefficient in Equation II.
* Elimination; a coefficient in Equation II is an integer multiple of a coefficient in Equation I.
Step-by-step explanation:
Equation I: 4x − 5y = 4
Equation II: 2x + 3y = 2
These equation can only be solved by Elimination method
Where to Eliminate x :
We Multiply Equation I by a coefficient of x in Equation II and Equation II by the coefficient of x in Equation I
Hence:
Equation I: 4x − 5y = 4 × 2
Equation II: 2x + 3y = 2 × 4
8x - 10y = 20
8x +12y = 6
Therefore, the valid reason using the given solution method to solve the system of equations shown is:
* Elimination; a coefficient in Equation I is an integer multiple of a coefficient in Equation II.
* Elimination; a coefficient in Equation II is an integer multiple of a coefficient in Equation I.
Answer:
-32-p
Step-by-step explanation:
you combine the like terms which would be -21 and -11. that gives you -32 and -p is just by its self since there are no terms to combine it with
Answer:
answer is D) 0, -5
Step-by-step explanation:
