ans is b...multiply eqn 1 by -2 and eqn 2 by 3
=》 -6x + 10y = 4...eqn 3
=》 6x + 3y = 9...eqn 4
add eqn 3 and 4...
=》13y = 13
=》y =1 and x = 1
Answer:
A: Right
B: Acute
C: Obtuse
D: Acute
Step-by-step explanation:
A: Right
(90 degrees)
B: Acute
(less than 90)
C: Obtuse
(over 90 degrees)
D: Acute
(less than 90)
Answer:
That seems a little hard-
Step-by-step explanation:
Factor -20x+32
which comes out to 4(-5x+8)
answer: 4(-5x+8)
hope this helped :)
<h2>>>> Answer <<<</h2>
Let's check which polynomial is divisible by ( x - 1 ) using hit , trial and error method .
A ( x ) = 3x³ + 2x² - x
The word " divisible " itself says that " it is a factor "
Using factor theorem ;
Let;
=> x - 1 = 0
=> x = 1
Substitute the value of x in p ( x )
p ( 1 ) =
3 ( 1 )³ + 2 ( 1 )² - 1
3 ( 1 ) + 2 ( 1 ) - 1
3 + 2 - 1
5 - 1
4
This implies ;
A ( x ) is not divisible by ( x - 1 )
Similarly,
B ( x ) = 5x³ - 4x² - x
B ( 1 ) =
5 ( 1 )³ - 4 ( 1 )² - 1
5 ( 1 ) - 4 ( 1 ) - 1
5 - 4 - 1
5 - 5
0
This implies ;
B ( x ) is divisible by ( x - 1 )
Similarly,
C ( x ) = 2x³ - 3x² + 2x - 1
C ( 1 ) =
2 ( 1 )³ - 3 ( 1 )² + 2 ( 1 ) - 1
2 ( 1 ) - 3 ( 1 ) + 2 - 1
2 - 3 + 2 - 1
4 - 4
0
This implies ;
C ( x ) is divisible by ( x - 1 )
Similarly,
D ( x ) = x³ + 2x² + 3x + 2
D ( 1 ) =
( 1 )³ + 2 ( 1 )² + 3 ( 1 ) + 2
1 + 2 + 3 + 2
8
This implies ;
D ( x ) is not divisible by ( x - 1 )
<h2>Therefore ; </h2>
<h3>B ( x ) & C ( x ) are divisible by ( x - 1 ) </h3>
