The solution to the given equation is x = ±7. The correct option is d. x = ±7
<h3>Solving quadratic equations </h3>
From the question, we are to determine the solution to the given equation
The given equation is
x²-49 = 0
NOTE: I assumed the variable is x
Solving the equation
x²-49 = 0
Writing in the form ax²= c
x² = 49
∴ x = ±√49
x = ±7
Hence, the solution to the given equation is x = ±7
Learn more on Solving quadratic equations here: brainly.com/question/24334139
#SPJ1
(2x-5y)+(x+y)=(2x+x)+(-5y+y)=3x-4y.
therefore:
(2x-5y)+(x+y)=3x-4y
Twelve and one hundred sixty-five thousandths
Answer:
There are no solutions
Step-by-step explanation:
#1. B
<span>(z * z^2 + z * 2z + z * 4) – (-2 *z^2 – (-2) 2z – (-2) 4)
Z^3 + 2z^2 + 4z – 2z^2 -4z – 8
Z^3 + 2z^2 – 2z^2 + 4z – 4z – 8
Z^3 - 8
</span>
#2 and #3. D
<span>(x + y)(x + 2)
x^2 + 2x + yx + 2y
</span>
#4. D.
<span>(x - 7)(x + 7)(x- 2)
x^2 + 7x – 7x -49
x^2 + x – 49
x^2 -49
(x^2 – 49 ) (x – 2)
x^3 – 2x^2 – 49x + 98
</span>
#5. C
(y - 4) = 0
y = 4
(x + 3)= 0
x = -3
#6. A and B