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2 years ago

find an equation involving g, h, and k that makes this augmented matrix correspond to a consistent system:

Mathematics

1 answer:

madam [21]2 years ago

3 0

The equation involving g, h, and k is mathematically given as

0=b ( "b" is equal to zero)

<h3>What is the equation involving g, h, and k that makes this augmented matrix correspond to a consistent system?</h3>

Generally, the equation for is mathematically given as

$\begin{aligned}&{\left[\begin{array}{cccc}1 & -4 & 7 & g \\0 & 3 & -5 & h \\-2 & 5 & -9 & k\end{array}\right]} \\\\&\sim\left[\begin{array}{cccc}1 & -4 & 7 & g \\0 & 3 & -5 & h \\-2 & 5 & -9 & k\end{array}\right] \\\\&\sim\left[\begin{array}{cccc}1 & -4 & 7 & g \\0 & 3 & -5 & h \\0 & -3 & 5 & 2 g+k\\\\\end{array}\right] \quad R_{3}+2 R_{1} \\\\&\sim\left[\begin{array}{cccc}1 & -4 & 7 & g \\0 & 3 & -5 & h \\0 & 0 & 0 & 2 g+k+h\end{array}\right] \quad R_{3}+R_{2}\end{aligned}$

In conclusion, Let "b" symbolize the integer 2g+k+h The final equation that is represented by the augmented matrix that was just given is thus 0=b.

Because this equation can only be satisfied if and only if the variable "b" is equal to zero, the first system can only have a solution if and only if the combination of g, k, and h equals 0.

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Trent is building a new L-shaped desk for his room. He wants to make sure it

babymother [125]

Answer: 33 square feet

Step-by-step explanation:

First, let's take the area of the rectangle shown here which is 6ft long and 4ft wide

6x4=24

Now let's take the area of the square which is 3ft long and 3ft wide

3x3=9

Now add 24 and 9 for the total area

24+9=33 square feet

4 0

3 years ago

Write a definite integral that represents the area of the region. (Do not evaluate the integral.) y1 = x2 + 2x + 3 y2 = 2x + 12F

Svet_ta [14]

Answer:

$A = \int\limits^3__-3}{9}-{x^{2}} \, dx = 36$

Step-by-step explanation:

The equations are:

$y = x^{2} + 2x + 3$

$y = 2x + 12$

The two graphs intersect when:

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

$x^{2} = 0$

$x_{1} = 3\\x_{2} = -3$

To find the area under the curve for the first equation:

$A_{1} = \int\limits^3__-3}{x^{2} + 2x + 3} \, dx$

To find the area under the curve for the second equation:

$A_{2} = \int\limits^3__-3}{2x + 12} \, dx$

To find the total area:

$A = A_{2} -A_{1} = \int\limits^3__-3}{2x + 12} \, dx -\int\limits^3__-3}{x^{2} + 2x + 3} \, dx$

Simplifying the equation:

$A = \int\limits^3__-3}{2x + 12}-({x^{2} + 2x + 3}) \, dx = \int\limits^3__-3}{9}-{x^{2}} \, dx$

Note: The reason the area is equal to the area two minus area one is that the line, area 2, is above the region of interest (see image).

3 0

3 years ago

Give the equation of the line perpendicular to y +7= -2(x - 6) through the point (6,-3)

lilavasa [31]

Note: this answer assumes the equation of the line can be put in slope-intercept form.

Answer:

$y + 3 = \frac{1}{2}(x-6)$

Step-by-step explanation:

1) First, find the slope of y + 7 = -2 (x - 6). We can see that it's already in point-slope form, or $y-y_1 = m (x-x_1)$ format. Remember that the number in place of the $m$ is the slope. Therefore, -2 is the slope of that equation.

What we need is the slope that is perpendicular to that, though. So, find the opposite reciprocal of -2. To do this, change its sign, convert it into a fraction ( $-\frac{2}{1}$ ), and flip its numerators and denominators. Therefore, the perpendicular slope would be $\frac{1}{2}$ .

2) Now that we have a slope and a point the line passes through, we can write an equation using the point-slope formula, $y-y_1 = m (x-x_1)$ . In order to write an equation, the $x_1$ , $y_1$ , and $m$ have to be substitute for with real values.

The $m$ represents the slope. We already calculated that in the last step, so put $\frac{1}{2}$ in place of the $m$ . The $x_1$ and $y_1$ represent the x and y values of a point the line passes through. We know that the line has to pass through (6, -3), so substitute 6 for $x_1$ and -3 for $y_1$ :

$y - (-3) = \frac{1}{2}(x -(6))\\y + 3 = \frac{1}{2}(x - 6)$