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3 years ago

Determine the x- and y-intercepts of the graph of y = 1/4x - 2

Mathematics

2 answers:

Olegator [25]3 years ago

8 0

8,-2. place the x as 8 and y as -2.

EleoNora [17]3 years ago

5 0

Answer:

The y-intecept is $(0,-2)$ , and the x-intercept is $(8,0)$ .

Step-by-step explanation:

The interception points with the axes can be foun by making zero one variable.

If you wanna find y-interception, we must make x = 0.

If you wann find x-interception, we must make y = 0.

The given function is

$y=\frac{1}{4}x-2$

For $x=0$ , we have

$y=\frac{1}{4}(0)-2\\y=-2$

For $y=0$ , we have

$y=\frac{1}{4}x-2\\0=\frac{1}{4}x-2\\2=\frac{1}{4}x\\x=8$

Therefore, the y-intecept is $(0,-2)$ , and the x-intercept is $(8,0)$ .

The image attached show this.

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2 years ago

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What is the value of the expression 18-2x(4+3)

Yanka [14]

Answer:

12-8x

Step-by-step explanation:

18-2x(4+3) =

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3 0

3 years ago

Solve system of equations: 4l + 8m = 40 5l + 2m = 28 Please show steps!

ziro4ka [17]

Answer:

l = 9/2 or 4.5

m = 11/4 or 2.75

Step-by-step explanation:

4l + 8m = 40

8m = 40 - 4I

m = 5 - 0.5l

5l + 2m = 28

5l + 2(5 - 0.5l) = 28

5l + (10 - l) = 28

4l + 10 = 28

4l = 18

l = 18/4 or 9/2

l = 4.5

4l + 8m = 40

4(4.5) + 8m = 40

18 + 8m = 40

8m + 18 = 40

8m = 22

m = 22/8

m = 2.75

3 0

2 years ago

Find the directional derivative of the function at the given point in the direction of the vector v. G(r, s) = tan−1(rs), (1, 3)

alexandr1967 [171]

The directional derivative of $f$ at the given point in the direction indicated is $\frac{5}{2}$ .

<h3>How to calculate the directional derivative of a multivariate function</h3>

The directional derivative is represented by the following formula:

$\nabla_{\vec v} f = \nabla f (r_{o}, s_{o})\cdot \vec v$ (1)

Where:

$\nabla f (r_{o}, s_{o})$ - Gradient evaluated at the point $(r_{o}, s_{o})$ .
$\vec v$ - Directional vector.

The gradient of $f$ is calculated below:

$\nabla f (r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{\partial f}{\partial r}(r_{o},s_{o}) \\\frac{\partial f}{\partial s}(r_{o},s_{o}) \end{array}\right]$ (2)

Where $\frac{\partial f}{\partial r}$ and $\frac{\partial f}{\partial s}$ are the partial derivatives with respect to $r$ and $s$ , respectively.

If we know that $(r_{o}, s_{o}) = (1, 3)$ , then the gradient is:

$\nabla f(r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{s}{1+r^{2}\cdot s^{2}} \\\frac{r}{1+r^{2}\cdot s^{2}}\end{array}\right]$

$\nabla f (r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{3}{1+1^{2}\cdot 3^{2}} \\\frac{1}{1+1^{2}\cdot 3^{2}} \end{array}\right]$

$\nabla f (r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{3}{10} \\\frac{1}{10} \end{array}\right]$

If we know that $\vec v = 5\,\hat{i} + 10\,\hat{j}$ , then the directional derivative is:

$\nabla_{\vec v} f = \left[\begin{array}{cc}\frac{3}{10} \\\frac{1}{10} \end{array}\right] \cdot \left[\begin{array}{cc}5\\10\end{array}\right]$

$\nabla _{\vec v} f (r_{o}, s_{o}) = \frac{5}{2}$

The directional derivative of $f$ at the given point in the direction indicated is $\frac{5}{2}$ . $\blacksquare$

To learn more on directional derivative, we kindly invite to check this verified question: brainly.com/question/9964491

3 0

2 years ago

Write the slope intercept form of the equation of each line given the slope and y intercept

drek231 [11]

Answer:

the slope intercept form would look like this: y=6/5x+(-2)

Step-by-step explanation:

How to write slope-intercept form:

y=(slope)x+(y-intercept)

If you have the y-intercept and the slope, you can figure out this form.

Hope this helped!! :)

4 0

3 years ago