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2 years ago

X^2-14x-4=0, solve by completing the square

2 answers:

4 0

The answer is: x = 7 - √53 or x = 7 + √53

The general quadratic equation is: ax² + bx + c = 0.
But, by completing the square we turn it into: a(x + d)² + e = 0, where:
d = b/2a
e = c - b²/4a
Our quadratic equation is x² - 14x -4 = 0, which is after rearrangement:
So, a = 1, b = -14, c = -4
Let's first calculate d and e:
d = b/2a = -14/2*1 = -14/2 = -7
e = c - b²/4a = -4 - (-14)²/4*1 = -4 - 196/4 = -4 - 49 = -53
By completing the square we have:
a(x + d)² + e = 0

1(x + (-7))² + (-53) = 0
(x - 7)² - 53 = 0
(x - 7)² = 53
x - 7 = +/-√53
x = 7 +/- √53

Therefore, the solutions are:
x = 7 - √53
or
x = 7 + √53

devlian [24]2 years ago

4 0

×^2-14×+49-49-4=0
(×-7)^2-53=0
(×-7-7.3)(×-7+7.3)
(×-14.3)(×+0.3)=0

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Please help!! Write a matrix representing the system of equations

frozen [14]

Answer:

$(4, -1, 3)$

Step-by-step explanation:

We have the system of equations:

$\left\{ \begin{array}{ll} x+2y+z =5 \\ 2x-y+2z=15\\3x+y-z=8 \end{array} \right.$

We can convert this to a matrix. In order to convert a triple system of equations to matrix, we can use the following format:

$\begin{bmatrix}x_1& y_1& z_1&c_1\\x_2 & y_2 & z_2&c_2\\x_3&y_2&z_3&c_3 \end{bmatrix}$

Importantly, make sure the coefficients of each variable align vertically, and that each equation aligns horizontally.

In order to solve this matrix and the system, we will have to convert this to the reduced row-echelon form, namely:

$\begin{bmatrix}1 & 0& 0&x\\0 & 1 & 0&y\\0&0&1&z \end{bmatrix}$

Where the (x, y, z) is our solution set.

Reducing:

With our system, we will have the following matrix:

$\begin{bmatrix}1 & 2& 1&5\\2 & -1 & 2&15\\3&1&-1&8 \end{bmatrix}$

What we should begin by doing is too see how we can change each row to the reduced-form.

Notice that R₁ and R₂ are rather similar. In fact, we can cancel out the 1s in R₂. To do so, we can add R₂ to -2(R₁). This gives us:

$\begin{bmatrix}1 & 2& 1&5\\2+(-2) & -1+(-4) & 2+(-2)&15+(-10) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\0 & -5 & 0&5 \\3&1&-1&8 \end{bmatrix}$

Now, we can multiply R₂ by -1/5. This yields:

$\begin{bmatrix}1 & 2& 1&5\\ -\frac{1}{5}(0) & -\frac{1}{5}(-5) & -\frac{1}{5}(0)& -\frac{1}{5}(5) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3&1&-1&8 \end{bmatrix}$

From here, we can eliminate the 3 in R₃ by adding it to -3(R₁). This yields:

$\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3+(-3)&1+(-6)&-1+(-3)&8+(-15) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&-5&-4&-7 \end{bmatrix}$

We can eliminate the -5 in R₃ by adding 5(R₂). This yields:

$\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0+(0)&-5+(5)&-4+(0)&-7+(-5) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&-4&-12 \end{bmatrix}$

We can now reduce R₃ by multiply it by -1/4:

$\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\ -\frac{1}{4}(0)&-\frac{1}{4}(0)&-\frac{1}{4}(-4)&-\frac{1}{4}(-12) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}$

Finally, we just have to reduce R₁. Let's eliminate the 2 first. We can do that by adding -2(R₂). So:

$\begin{bmatrix}1+(0) & 2+(-2)& 1+(0)&5+(-(-2))\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 1&7\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}$

And finally, we can eliminate the second 1 by adding -(R₃):

$\begin{bmatrix}1 +(0)& 0+(0)& 1+(-1)&7+(-3)\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 0&4\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}$

Therefore, our solution set is $(4, -1, 3)$

And we're done!

3 0

3 years ago

For the equation : y = 2x + 10. Write any two possible solutions in ordered form. (Hint: e.G ( 4, 18))

nalin [4]

Answer:

(0, 10)

and

(1, 12)

5 0

3 years ago

Read 2 more answers

Use these functions to answer the questions.

Elenna [48]

Part A
We have $f\left(x\right)=\left(x+3\right)^2-1$ . To solve for the x-intercept, we set f(x) equal to 0. That is
$\left(x+3\right)^2-1=0$

$\left(x+3\right)^2=1$

Take the square root of both sides,
$x+3=1$

$x=-2$

The x-intercept is (-2,0).

To solve for the y-intercept, we set x=0. That is
$y=\left(0+3\right)^2-1=3^2-1=9-1=8$

The y-intercept is (0, 8)

The coordinates of the optimum point are actually the vertex which can be easily seen from the vertex form equation given above. The minimum point is (-3, -1).

Part B.
We have $g\left(x\right)=-2x^2+8x+3$ .
Factor out -2
$=-2\left(x^2-4x\right)+3$

Complete the square
$=-2\left(x^2-4x+4\right)+3-2\left(4\right)$

Simplify
$g(x)=-2\left(x-2\right)^2-5$

Part C
We have $h\left(t\right)=-16t^2+28t$ .
The maximum height is 12.25 feet after 0.875 seconds from the time of the jump. The dolphin will be back in the water after 1.75 seconds. The graph of the jump is shown in the photo.

4 0

3 years ago

In a recent school newspaper survey, 3,000 randomly selected teenagers were asked to cite their primary transportation method to

Verizon [17]

Answer: There is a 90% chance that the true proportion of teenagers who drive their own car to school will lie in (0.5907, 0.9093).

Step-by-step explanation:

Interpretation of a% confidence interval : A person can be a% confident that the true population parameter lies in it.

Here, A 90% confidence interval to estimate the true proportion of teenagers who drive their own car to school is found to be (0.5907, 0.9093).

i.e. A person can be 90% confident that the true proportion of teenagers who drive their own car to school lies in (0.5907, 0.9093).

Hence, correct interpretation is : There is a 90% chance that the true proportion of teenagers who drive their own car to school will lie in (0.5907, 0.9093).

7 0

2 years ago

Could someone help

zhenek [66]

Answer: 3v^2w^2 + 10v^2w^3 (or the first option on the screen)

7 0

2 years ago

Read 2 more answers