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.
To learn more about real roots refer the below link
brainly.com/question/24147137
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Answer:
where is question ???
Step-by-step explanation:
Answer:
$4.06
Step-by-step explanation:
23 / 5 = 4
Remainder = 3
$3 into dimes = 30 dimes
30 / 5 = 6
$4 + €6 = $4.06
Answer:
<h2>6÷4(7+8) </h2><h2>= 22,5</h2>
<h2>8-6(4+9)</h2><h2>= 26</h2>
<h2>4÷8(9+3)</h2><h2>= 6</h2>
Step-by-step explanation:
<h2>6÷4(7+8)</h2><h3>= 6÷4(15)</h3><h3>= 1,5 × 15</h3><h3>= 22,5</h3>
<h2>8-6(4+9)</h2><h3>= 8-6(13)</h3><h3>= 2 × 13</h3><h3>= 26 </h3>
<h2>4÷8(9+3)</h2><h3>= 4÷8(12)</h3><h3>= 0,5 × 12</h3><h3>= 6</h3>
A polygon with four sides is known as a quadrilateral.
Quadrilaterals have four sides and four angles as well.
There's many different types of quadrilaterals:
square
rectangle
rhombus
trapezoid
The trapezoid has to meet qualifications to be considered a quadrilateral, such as a pair of parellel lines.
