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3 years ago

Graph the lines by finding the points of intersection with the axes (intercepts):

Mathematics

2 answers:

Ostrovityanka [42]3 years ago

5 0

To solve this problem you must apply the proccedure shown below:

1. You have the following function:

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

2. You can rewrite it as following:

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

3. The line intersects the y-axis when $x=0$ , therefore:

$y=\frac{(0+1)}{2}=\frac{1}{2} =0.5$

4. The line intersects the x-axis when $y=0$ :

$0=\frac{(-x+1)}{2}\\ 0=-x+1\\ x=1$

5. Now plot the points (0, 0.5) and (1,0) in a graph , as you can see in the figure attached.

Therefore, the answer is: The points of intersection are (0, 0.5) and (1,0).

rewona [7]3 years ago

5 0

Answer:

Graph in attachment.

Step-by-step explanation:

Given: $y=-\dfrac{1}{2}x-1$

To graph the line using intercepts with axis.

x-intercept and y-intercept

For x-intercept, Put y=0 into equation

$0=-\dfrac{1}{2}x-1$

$x=-2$

x-intercept: (-2,0)

For y-intercept, Put x=0 into equation

$y=-\dfrac{1}{2}\cdot 0-1$

$y=-1$

y-intercept: (0,-1)

Now draw the graph by plotting the points.

Please have a look with attachment for graph.

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In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 3.0 m/s up a 22.0° inclined track

Aloiza [94]

Answer:

Approximately $0.31\; \rm m$ , assuming that $g = 9.81\; \rm N \cdot kg^{-1}$ .

Step-by-step explanation:

Initial kinetic energy of the sled and its passenger:

$\begin{aligned}\text{KE} &= \frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 14\; \rm kg \times (3.0\; \rm m\cdot s^{-1})^{2} \\ &= 63\; \rm J\end{aligned}$ .

Weight of the slide:

$\begin{aligned}W &= m \cdot g \\ &= 14\; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \\ &\approx 137\; \rm N\end{aligned}$ .

Normal force between the sled and the slope:

$\begin{aligned}F_{\rm N} &= W\cdot \cos(22^{\circ}) \\ &\approx 137\; \rm N \times \cos(22^{\circ}) \\ &\approx 127\; \rm N\end{aligned}$ .

Calculate the kinetic friction between the sled and the slope:

$\begin{aligned} f &= \mu_{k} \cdot F_{\rm N} \\ &\approx 0.20\times 127\; \rm N \\ &\approx 25.5\; \rm N\end{aligned}$ .

Assume that the sled and its passenger has reached a height of $h$ meters relative to the base of the slope.

Gain in gravitational potential energy:

$\begin{aligned}\text{GPE} &= m \cdot g \cdot (h\; {\rm m}) \\ &\approx 14\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times h\; {\rm m} \\ & \approx (137\, h)\; {\rm J} \end{aligned}$ .

Distance travelled along the slope:

$\begin{aligned}x &= \frac{h}{\sin(22^{\circ})} \\ &\approx \frac{h\; \rm m}{0.375}\end{aligned}$ .

The energy lost to friction (same as the opposite of the amount of work that friction did on this sled) would be:

$\begin{aligned} & - (-x)\, f \\ = \; & x \cdot f \\ \approx \; & \frac{h\; {\rm m}}{0.375}\times 25.5\; {\rm N} \\ \approx\; & (68.1\, h)\; {\rm J}\end{aligned}$ .

In other words, the sled and its passenger would have lost (approximately) $((137 + 68.1)\, h)\; {\rm J}$ of energy when it is at a height of $h\; {\rm m}$ .

The initial amount of energy that the sled and its passenger possessed was $\text{KE} = 63\; {\rm J}$ . All that much energy would have been converted when the sled is at its maximum height. Therefore, when $h\; {\rm m}$ is the maximum height of the sled, the following equation would hold.