The equation y= 2 has one real root and that is x=-1.
What is real roots of the equation?
We are aware that when we resolve a linear or quadratic equation, we always arrive at the value variable of the equation, or, to put it another way, we always locate the equation's solution. This "solution" is what we refer to as the real roots. For instance, when the equation -7x+12=0 is solved, the actual roots are 3 and 4.
Here given,
=> y = 2
Take y=0 then,
=> 2=0
=> =0
=>(x+1)=0
=> x=-1
Hence the given equation has one real root and that is x=-1.
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A quadrilateral PQRS is a parallelogram. PQ and RS are opposite sides then the answer of this particular question is , x = 4, the parallelogram PQRS is a rhombus, QR = 31, The opposite angles are congruent. Thus, more than one answer is correct.
According to the question,
PQ = 7x+12; RS = 5x -20 and QR = 31
Solving PQ and RS we get x = 4. Thus, PQ = 40 and RS = 40.
A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent. A quadrilateral that has all sides equal and opposite sides parallel is called Rhombus.
Thus, the parallelogram PQRS is a rhombus.
Hence, a quadrilateral PQRS is a parallelogram. PQ and RS are opposite sides then the answer of this particular question is , x = 4, the parallelogram PQRS is a rhombus, QR = 31, The opposite angles are congruent. Thus, more than one answer is correct.
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Answer:
No
Step-by-step explanation:
The definition of a prime number is a number that only has factors of 1 and itself. So there are only 2 factors.
4 has 3 factors: 1, 2, and 4. It has factors of 1 and itself (4), but also has an extra factor (2), so 4 is not a prime number
Step-by-step explanation:
The slope of the above line = -2
slope of the line perpendicular to the above line = 1/2
If it passes through point (4,-3) , it's equation
A = 1, B = -2 and C = -10
A+B+C= 1+(-2)+(-10)
= 1-2-10
= 1-12
= -11