Answer:
The correct option is x=3 , y=2
Step-by-step explanation:
According to the HL theorem if the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and one leg of another right triangle then the triangles are congruent.
By using this theorem we can set up the system of equations as follows:
x=y+1 ...(1)
2x+3= 3y + 3 ..(2)
Now we will plug the value y+1 of equation 1 in equation 2.
2x+3 = 3y+3
2(y+1)+3=3y+3
2y+2+3=3y+3
2y+5=3y+3
Now combine the like terms:
5-3=3y-2y
2=y
y=2
Now plug the value y=2 in equation 1.
x=y+1
x=2+1
x=3
Thus the values of(x.y) are{(2,3)}
Therefore the correct option is x=3 , y=2 .....
Answer:
The answer is D
Step-by-step explanation:
I did it on edg
Answer:
<em>64 ; 8</em>
Step-by-step explanation:
a² ± 2ab + b² = (a ± b)²
x² + 2(8)(x) + 8² =
x² + 16x + <em>64</em> =
(x + <em>8</em>)²
Answer:
We show that f(x) n+8/6n = 6 x n = 0
which flips the n+8/1 = 0+8/0-6= x = 3 this is the range.
For the HA we would work left to right.
x goes to positive or negative infinity and is determined by the highest degree terms of the polynomials in the numerator and the denominator. This particular function has polynomials of degree 0 in both the numerator and the denominator
If say n+8 was n+2 then we would use the 2/-2+3 and get 1 and show the hole as the source;
hole : -2+1 as non equal sign. but not in the case of n+8/6n
-2+1 represents 1/3 symmetry.
We see for n+8/6n with interpreted back into the zero format minus
-0+8/-0-6 we see there is symmetry and can work on the left side of graph and flip over. Where 0 = n+8 and 1=nx6
Step-by-step explanation:
There would be no way of doing the others unless the exponents had been squared ^2
If they were squared then the domain will be (-infinity -3) parenthesis
union of( -3 -2) union of +2 to negative infinity.
There is not a vertical asymptote as the numerator divides into dominator at point 8 as a decimal.
The holes are then closed.
