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Airida [17]
2 years ago
13

What are the solution(s) to the quadratic equation 50 – x2 = 0?

Mathematics
2 answers:
Novay_Z [31]2 years ago
8 0

Answer:

Solution of the given Quadratic equation are +5√2 and -5√2.

 

Step-by-step explanation:

We are given Quadratic Equation,

50-x^2=0

To find: Solutions of the given quadratic equation.

Consider,

50-x^2=0

transpose x² to RHS,

50=x^2

Now transpose whole RHS to LHS and whole LHS to RHS,

x^2=50

Now, Apply square root both sides,

x=\pm\sqrt{50}

x=\pm\sqrt{2\times5\times5}

x=\pm5\sqrt{2}

Therefore, Solution of the given Quadratic equation are +5√2 and -5√2.

Alecsey [184]2 years ago
5 0
I hope this helps you



x^2=50


x= 5 square root of 2
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Solve the system using elimination. x + 2y = –6 3x + 8y = –20 (–4, –1) (–4, 4) (–1, –4) (3, 1)
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Here is the histogram of a data distribution. All class widths are 1. What is the median of this distribution?
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Answer:

The median is 4.

Step-by-step explanation

if you layout all the number, you get

1,2,2,3,3,3,4,5,5,5,6,6,7

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(Ex. 3,4,5,7; 4.5 is the answer)

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2 years ago
1. Contaminated water is subjected to a cleaning process. The concentration of the pollutants is initially 5 mg per liter of wat
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Answer:

C(t)=5\cdot(0.9)^t

Step-by-step explanation:

The exponential function is often used to model natural growing or decaying processes, where the change is proportional to the actual quantity.

An exponential decaying function is expressed as:

C(t)=C_o\cdot(1-r)^t

Where:

C(t) is the actual value of the function at time t

Co is the initial value of C at t=0

r is the decaying rate, expressed in decimal

The concentration of the pollutants starts at Co=5 mg/lt. We also know the pollutant reduces its concentration by 10% each hour. This gives us a value of r = 10% / 100 = 0.1

Substituting into the general equation:

C(t)=5\cdot(1-0.1)^t

Operating:

\boxed{C(t)=5\cdot(0.9)^t}

7 0
2 years ago
Using V = lwh , find an expression in factored form for the volume of this prism. Explain the general steps you used to arrive a
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Find the real numbers x and y if -3+ix^2y and x^2+y+4i are conjugate of each other. Pls solve with the steps
Firdavs [7]
ANSWER
x = ±1 and y = -4.
Either x = +1 or x = -1 will work

EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same

For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y

For x² + y + 4i
⇒ real part: x² + y (since x, y are real numbers)
⇒ imaginary part: 4

Therefore, for the two expressions to be conjugates, we must satisfy the two conditions. 

Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the 

   x²y = -4 ... (I)

Condition 2: Real parts are the same

   x² + y = -3 ... (II)

We have a system of equations since both conditions must be satisfied

   x²y = -4 ... (I)
   x² + y = -3 ... (II)

We can rearrange equation (II) so that we have

   y = -3 - x² ... (II)

Substituting into equation (I)

   x²y = -4 ... (I)
   x²(-3 - x²) = -4
   -3x² - x⁴ = -4
   x⁴ + 3x² - 4 = 0
   (x² + 4)(x² - 1) = 0
   (x² + 4)(x-1)(x+1) = 0

Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.

Solve for y:

   y = -3 - x² ... (II)
   y = -3 - (±1)²
   y = -3 - 1
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So x = ±1 and y = -4. We can confirm this results in conjugates by substituting into the expressions:

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   = -3 - 4i

   x² + y + 4i
   = (±1)² - 4 + 4i
   = 1 - 4 + 4i
   = -3 + 4i

They result in conjugates
4 0
2 years ago
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