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3 years ago

How do the arrows for +1 and -1 show both distance and direction

Mathematics

2 answers:

daser333 [38]3 years ago

8 0

Ur going up +1 and down -1

Taya2010 [7]3 years ago

5 0

Negative one is to the left of the graph and positive is to the right. Example: -1<------>+1

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A hot air ballon ascended at the rate of 400 dreams per hot for 3 hours and 15 min. What was the total distance that the ballon

alexandr1967 [171]

If it is 400 m/s then it's the answer

v= s/t
400m/s= s/11700 195×60=11,700 sec
s= 400 × 11700
s= 4,680,000 m/s ^2

11,700×400=4,680,000

4 0

3 years ago

Can someone help me out with my math homework my grades are ba

kupik [55]

<h3>Answer: 18</h3>

==========================================================

Explanation:

2 cm on paper (or on screen) represents 3 meters in real life.

The backyard is 2 cm horizontally across on the screen, so it's 3 meters across in real life. This is the shorter dimension.

The longer dimension is 6 meters. We can think of it like this

2 cm : 3 meters

2*2 cm : 2*3 meters .... multiply both sides by 2

4 cm : 6 meters

The real backyard is 6 meters by 3 meters to give an area of 6*3 = 18 square meters.

7 0

3 years ago

The circumference C of a circle with radius r can be calculated using the formula C=2pir. Which formula represents r in terms of

DaniilM [7]

Answer:

Step-by-step explanation:

If C=2πr

r=C/2π (change of sbject)

6 0

3 years ago

Solve 6x+3y=−1 for y

vlabodo [156]

3y=-6x-1
y=-2x-1/3 is the answer

8 0

3 years ago

Read 2 more answers

The velocity of a turtle is recorded at 1 minute intervals (in meters per second). Use the right-endpoint approximation to estim

aliina [53]

Answer:

The total distance that the turtle traveled during the 5 seconds recorded is $Distance \:traveled\approx 3.838\:\frac{m}{s}$

Step-by-step explanation:

To estimate distance traveled of an object moving in a straight line over a period of time, from discrete data on the velocity of the object, we use a Riemann Sum. If we have a table of values

$\left\begin{array}{ccccccc}time\:=\:t_i&t_0=0&t_1&t_2&...&t_n\\velocity\:=\:v(t_i)&v(t_0)&v(t_1)&v(t_2)&...&v(t_n)\end{array}\right$

where $\Delta t=t_i-t_{i-1}$ , then we can approximate the displacement on the interval $[t_{i-1},t_i]$ by $v(t_{i}) \times\Delta t$ .

Therefore the distance traveled of the object over the time interval $[0,t_n]$ can be approximated by

$Distance \:traveled\approx |v(t_1)|\Delta t+|v(t_2)|\Delta t+...+|v(t_n)|\Delta t$

This is the right endpoint approximation.

We are given a table of values for <em>v(t)</em>

$\left\begin{array}{cccccccc}t(sec)&0&1&2&3&4&5\\v(t)&0.078&0.83&0.75&0.98&0.853&0.425\end{array}\right$

Applying the right endpoint approximation formula we get,

$\Delta t = 1\sec$

$Distance \:traveled\approx |v(t_1)|\Delta t+|v(t_2)|\Delta t+|v(t_3)|\Delta t+|v(t_4)|\Delta t+|v(t_5)|\Delta t\\\\Distance \:traveled\approx 0.83(1)+0.75(1)+0.98(1)+0.853(1)+0.425(1)\\\\Distance \:traveled\approx 3.838\:\frac{m}{s}$

The total distance that the turtle traveled during the 5 seconds recorded is $Distance \:traveled\approx 3.838\:\frac{m}{s}$

4 0

3 years ago