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1 year ago

When constructing a perpendicular bisector why must the compass opening be greater than 1/2.

1 answer:

4 0

It is critical to open a compass with over half the way in order for the arcs formed to meet for perpendicular bisector.

<h3>What is perpendicular bisector?</h3>

A perpendicular bisector would be a line that cuts a line segment in half and forms a 90-degree angle at the intersection point. In other words, a perpendicular bisector separates a line segment now at midpoint, forming a 90-degree angle.

Now, consider an example;

When you wish to build a perpendicular line on a line segment, like line AB, you do the following;

Set the compass on a radius more than half the length of the line AB.
Using A as your center, draw an arc above and below line AB.
With the same radius and B as our center, draw additional arcs on top or below line AB to join a first arcs on the both side of the line.
Join the two arc intersections to cut line AB at M.

A line AB appears to be bisected perpendicularly as a result. The arcs would not have met if the compass is opened less than half way down line AB.

To know more about perpendicular bisector, here

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2. SALES TAX Olivia purchases a DVD movie priced

Nataly_w [17]

Answer:

$23.42

Step-by-step explanation:

To find the sales tax, you need to 6.5% of $21.99
To do this you have to times need to turn 6.5% into a decimal, which is 0.065 (i moved decimal point 2 numbers to the left)
then you do 0.065 × 21.99 which equals 1.42935
now you have to add 21.99 with 1.42935 = 23.41935
because of the context of the question, you need the answer needs to be rounded up to the nearest hundredth.
making the answer $23.42 total

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3 years ago

-5/2n-5 + 2n/2n-5. Step by step explanation??

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8 0

3 years ago

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.

$((137 + 68.1)\, h)\; {\rm J} = 63\; {\rm J}$ .

Solve for $h$ :

$(137 + 68.1)\, h = 63$ .

$\begin{aligned} h &= \frac{63}{137 + 68.1} \approx 0.31\; \rm m\end{aligned}$ .

Therefore, the maximum height that this sled would reach would be approximately $0.31\; \rm m$ .

7 0

2 years ago

Y=2x+5 y=3x−1 solve by subsitution

loris [4]

Answer:

(6,17)

Step-by-step explanation:

2x+5=3x-1

-3x -3x

-x+5=-1

-5 -5

-x=-6

^divided by -1 on both sides

x=6

y=2x+5

y=(2)(6)+5

y=17

5 0

2 years ago

What is the area of a square computer screen with 21 inch sides?

notka56 [123]

Side of the square computer screen = 21 inches
It is important to minutely read the details that are already given in the question. the most important information that is given in the question is that the computer screen ia square. Based on this information the answer to the question can be easily deduced.
Then
Area of the computer screen = (Side)^2
= (21)^2 inches square
= 441 inches^2
So the area of the square computer screen is 441 square inches. i hope the procedure is clear enough for you to understand.

6 0

3 years ago