Use the data to find the approximate values of the five-number summary. Select all that are true.

How many millimeters are in 1 meter

melisa1 [442]

The answer is 1000 millimeters in one meter

3 0

3 years ago

8x-3y times 5x5y over 10x6y2

erica [24]

Answer:

8x-5/4

Step-by-step explanation:

Cancel 5y x 5y/10 x 6y²=5/12y

8x-3y 5/12y

apply fraction rule a× b/c = a×b

____

c

8x-3y⁵/12y

8x-3y×5/12y

8x-3×5/12y

8x-3×5/12

=8x-5/4

6 0

2 years ago

Let g(x)=-5+2. Evaluate g(-1)

marta [7]

The answer would be g=3

8 0

4 years ago

HELP ME PLEASSEEEE use each numbers 2,3,7,8 and use operations make the the numbers FROM 20 TO 50 FOR EXAMPLE 2+3+7+8=20

lianna [129]

3•7=21 if times three times seven it equals twenty one

8 0

3 years ago

1. Consider an athlete running a 40-m dash. The position of the athlete is given by , where d is the position in meters and t is

sasho [114]

There is some information missing in the question, since we need to know what the position function is. The whole problem should look like this:

Consider an athlete running a 40-m dash. The position of the athlete is given by $d(t)=\frac{t^{2}}{6}+4t$ where d is the position in meters and t is the time elapsed, measured in seconds.

Compute the average velocity of the runner over the intervals:

(a) [1.95, 2.05]

(b) [1.995, 2.005]

(c) [1.9995, 2.0005]

(d) [2, 2.00001]

Answer

(a) 6.00041667m/s

(b) 6.00000417 m/s

(c) 6.00000004 m/s

(d) 6.00001 m/s

The instantaneous velocity of the athlete at t=2s is 6m/s

Step by step Explanation:

In order to find the average velocity on the given intervals, we will need to use the averate velocity formula:

$V_{average}=\frac{d(t_{2})-d(t_{1})}{t_{2}-t_{1}}$

so let's take the first interval:

(a) [1.95, 2.05]

$V_{average}=\frac{d(2.05)-d(1.95)}{2.05-1.95}$

we get that:

$d(1.95)=\frac{(1.95)^{3}}{6}+4(1.95)=9.0358125$

$d(2.05)=\frac{(2.05)^{3}}{6}+4(2.05)=9.635854167$

so:

$V_{average}=\frac{9.6358854167-9.0358125}{2.05-1.95}=6.00041667m/s$

(b) [1.995, 2.005]

$V_{average}=\frac{d(2.005)-d(1.995)}{2.005-1.995}$

we get that:

$d(1.995)=\frac{(1.995)^{3}}{6}+4(1.995)=9.30335831$

$d(2.005)=\frac{(2.005)^{3}}{6}+4(2.005)=9.363335835$

so:

$V_{average}=\frac{9.363335835-9.30335831}{2.005-1.995}=6.00000417m/s$

(c) [1.9995, 2.0005]

$V_{average}=\frac{d(2.0005)-d(1.9995)}{2.0005-1.9995}$

we get that:

$d(1.9995)=\frac{(1.9995)^{3}}{6}+4(1.9995)=9.33033358$

$d(2.0005)=\frac{(2.0005)^{3}}{6}+4(2.0005)=9.33633358$

so:

$V_{average}=\frac{9.33633358-9.33033358}{2.0005-1.9995}=6.00000004m/s$

(d) [2, 2.00001]

$V_{average}=\frac{d(2.00001)-d(2)}{2.00001-2}$

we get that:

$d(2)=\frac{(2)^{3}}{6}+4(2)=9.33333333$

$d(2.00001)=\frac{(2.00001)^{3}}{6}+4(2.00001)=9.33339333$

so:

$V_{average}=\frac{9.33339333-9.33333333}{2.00001-2}=6.00001m/s$

Since the closer the interval is to 2 the more it approaches to 6m/s, then the instantaneous velocity of the athlete at t=2s is 6m/s

8 0

3 years ago