Set up an equation
x+58=4x+37
21=3x
x=7
Jim:
x+58
7+58
$65
Jesse:
4x+37
4(7)+37
28+37
$65
Answer:
There are no extraneous solutions
Reasoning:
An extraneous solution is a solution that isn't valid, it might be imaginary like the square root of a negative number.
first we want to isolate z:
1+sqrt(z)=sqrt(z+5)
^2 all ^2 all
(1+sqrt(z))(1+sqrt(z))=z+5
expand
1+2sqrt(z)+z=z+5
-1 -z -z -1
2sqrt(z)=4
/2 /2
sqrt(z)=2
^2 all ^2 all
z=4
Since there is one solution and it is a real number, there are no extraneous solutions.
Answer:
Option (4)
Step-by-step explanation:
Given : XZ ≅ DF and ∠X ≅ ∠D
To prove : ΔXYZ ≅ ΔDEF
Statements Reasons
1). XZ ≅ DF 1). Given
2). ∠X ≅ ∠ D 2). Given
3). XY ≅ DE 3). Required information
4). ΔXYZ ≅ ΔDEF 4). By SAS property of congruence
Therefore, Option (4) will be the answer.
Step-by-step explanation:
So
WHOLE thing's rule (x,y)
(-1,6) one left 6 up
(1, -6) one right 6 down
(6,-1) six right one down
Good morning ☕️
______
Answer
-9+ 9і
___________________
Step-by-step explanation:
(-7+ 3i) - (2 - 6i) = -7+3i-2+6i = (-7-2)+(3i+6i) = -9+9i.
:)