The given equation is equivalent to Option D .
The mission option is
A. x+10= x+10
B. x+9 =x² + 20x + 100
C. x² +81 = x² + 100
D. x² + 81 = x² +20x + 100
<h3>What is an Equation ?</h3>
An equation is a mathematical statement formed when two algebraic expression is equated by an equal sign .
An equation is given in the question
![\rm \sqrt{x^2 + 81} = x +10](https://tex.z-dn.net/?f=%5Crm%20%5Csqrt%7Bx%5E2%20%2B%2081%7D%20%3D%20x%20%2B10)
![\rm {x^2 + 9^2} = (x +10)^2](https://tex.z-dn.net/?f=%5Crm%20%7Bx%5E2%20%2B%209%5E2%7D%20%3D%20%28x%20%2B10%29%5E2)
![\rm {x^2 + 81} = x^2 +20x + 100](https://tex.z-dn.net/?f=%5Crm%20%7Bx%5E2%20%2B%2081%7D%20%3D%20x%5E2%20%2B20x%20%2B%20100)
Therefore the given equation is equivalent to Option D .
To know more about Equation
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Answer:
Area = 16π
Step-by-step explanation:
Area of a circle = πr^2
= 64π
= sqrt64
r = 8 because 8 × 8 is 64
Radius = 8
Now, use the circumference equation:
= 2πr
= 2π(8)
= 16π
Opposite number is the same number with the opposite sign:
So the opposite of
![\frac{1}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B3%7D%20)
is
![-\frac{1}{3}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B3%7D%20)
, since the original number is positive
opposite of
![- \frac{7}{12}](https://tex.z-dn.net/?f=-%20%5Cfrac%7B7%7D%7B12%7D%20)
is
![\frac{7}{12}](https://tex.z-dn.net/?f=%5Cfrac%7B7%7D%7B12%7D%20)
, since the original number is negative
Answer:
D) 0 = 2(x + 5)(x + 3)
Step-by-step explanation:
Which of the following quadratic equations has no solution?
We have to solve the Quadratic equation for all the options in other to get a positive value as a solution for x.
A) 0 = −2(x − 5)2 + 3
0 = -2(x - 5) × 5
0 = (-2x + 10) × 5
0 = -10x + 50
10x = 50
x = 50/10
x = 5
Option A has a solution of 5
B) 0 = −2(x − 5)(x + 3)
Take each of the factors and equate them to zero
-2 = 0
= 0
x - 5 = 0
x = 5
x + 3 = 0
x = -3
Option B has a solution by one of its factors as a positive value of 5
C) 0 = 2(x − 5)2 + 3
0 = 2(x - 5) × 5
0 = (2x -10) × 5
0 = 10x -50
-10x = -50
x = -50/-10
x = 5
Option C has a solution of 5
D) 0 = 2(x + 5)(x + 3)
Take each of the factors and equate to zero
0 = 2
= 0
x + 5 = 0
x = -5
x + 3 = 0
x = -3
For option D, all the values of x are 0, or negative values of -5 and -3.
Therefore the Quadratic Equation for option D has no solution.