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Which equations are true for x = –2 and x = 2? Select two options x2 – 4 = 0 x2 = –4 3x2 + 12 = 0 4x2 = 16 2(x – 2)2 = 0

Svetllana [295]

The equation $x^{2} -4=0~~and~~4x^2=16$ are true for x = -2 and x = 2.

The given equations are given as:

$x^2-4=0\\\\x^2=-4\\\\3x^2+12=0\\\\4x^2=16\\\\2(x-2)^2=0$

We need to select two equations that are true for x = -2 and x = 2.

<h3>What are the solutions to an equation ?</h3>

The solutions of an equation are the values that satisfy the given equation or make the equation true when substituted for unknowns in the equations.

Example:

x - 2 = 2

Here the solution for x - 2 = 2 is 4 because x = 4 will make the equation true.

4 - 2 = 2

2 = 2

Let's find the solutions for each equation.

1.

$x^2 - 4=0$

$x^{2}$ = 4

x = $\sqrt{4}$ = ± 2

x = = 2 and x = -2

2.

$x^2 =-4$

x = $\sqrt{-4}$

x = $\sqrt{-1 \times 4}$ = $\sqrt{-1}\times\sqrt{4}$

x = ± 2 $\sqrt{-1}$

x = 2i and x = -2i where i = $\sqrt{-1}$

3.

3 $x^{2}$ + 12 = 0

$x^{2}$ = -12 / 3 = -4

x = $\sqrt{-1}$ $\sqrt{4}$

x = ± 2 $\sqrt{-1}$

x = 2i and x = -2i

4.

4 $x^{2}$ = 16

$x^{2}$ = 16 / 4

$x^{2}$ = 4

x = $\sqrt{4}$

x = ±2

x = 2 and x = -2

5.

$2(x-2)^2 = 0$

$(x-2)^2 = 0$

x - 2 = 0

x = 2.

Thus the equation $x^{2} -4=0~~and~~4x^2=16$ have solutions x = -2 and x = 2.

Learn more about solutions of equations here:

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4 0

2 years ago

How can x2 = 2x + 4 be set up as a system of equations?

mezya [45]

$\bf x^2=2x+4\implies 
\begin{cases}
y=x^2
\\\\
y=2x+4
\end{cases}\implies y=y$

3 0

4 years ago

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Mike used the computation StartFraction 7 Over (6 + 8 + 7) almost-equals 0.33 = 33 percent to find the percentage of eighth-grad

snow_tiger [21]

Answer: option B

Step-by-step explanation:

i took the test lol

3 0

3 years ago

A man wins in a gambling game if he gets two heads in five flips of a biased coin. the probability of getting a head with the co

Semenov [28]

The probability the man will win will be 13.23%. And the probability of winning if he wins by getting at least four heads in five flips will be 36.01%.

<h3>How to find that a given condition can be modeled by binomial distribution?</h3>

Binomial distributions consist of n independent Bernoulli trials.

Bernoulli trials are those trials that end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

P(X = x) = ⁿCₓ pˣ (1 - p)⁽ⁿ⁻ˣ⁾

A man wins in a gambling game if he gets two heads in five flips of a biased coin. the probability of getting a head with the coin is 0.7.

Then we have

p = 0.7

n = 5

Then the probability the man will win will be

P(X = 2) = ⁵C₂ (0.7)² (1 - 0.7)⁽⁵⁻²⁾

P(X = 2) = 10 x 0.49 x 0.027

P(X = 2) = 0.1323

P(X = 2) = 13.23%

Then the probability of winning if he wins by getting at least four heads in five flips will be

P(X = 4) = ⁵C₄ (0.7)⁴ (1 - 0.7)⁽⁵⁻⁴⁾

P(X = 4) = 5 x 0.2401 x 0.3

P(X = 4) = 0.3601

P(X = 4) = 36.01%

Learn more about binomial distribution here:

brainly.com/question/13609688

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4 0

2 years ago

A circular rug has a diameter of 6 feet. What is the area of the rug?

SVEN [57.7K]

B) 28.26 squared feet.

To fine area, you need the radius. radius is half of diameter. half of 6 is 3. A=πr2=π·32≈28.27433

3 0

3 years ago

Read 2 more answers