All categories

3 years ago

Displacement = 20 m north of starting line

Mathematics

1 answer:

valina [46]3 years ago

8 0

Answer:

move 20 m north of the starting line

Step-by-step explanation:

The runner will have a displacement of 20 m N of the starting line if that's where the runner stops when moving north from the starting line.

You might be interested in

Help plz..........................................................................................

Zarrin [17]

The answer is the last one

3 0

3 years ago

Read 2 more answers

1 point

SashulF [63]

Answer:

The length of the slide is 10 feet.

Step-by-step explanation:

To find the length of the slide you have to use the formula:

a^2 + b^2 = c^2 (^2 means squared)

In this case:

a= 6 and b=8 c= unknown

6 squared (6 times 6) is 36

8 squared (8 times 8) is 64

Add these two numbers to get:

36+64=100

6^2+8^2=100

However, the length of the slide in not 100 feet. In order to find the correct length you must find the square root of 100. Which is 10.

What number times itself equals 100? 10.

The length of the slide in 10 feet.

5 0

2 years ago

Help please number 7

chubhunter [2.5K]

The answer is D
Hope this helps

6 0

2 years ago

Please help asap 15 points that's more than usual just solve the question.

taurus [48]

Answer:

1/n^2 is the answer......

8 0

2 years ago

Apply Gaussian quadrature with n = 4 to approximate integrate sin x^2dx from 1 to 5

maks197457 [2]

First, recall that Gaussian quadrature is based around integrating a function over the interval [-1,1], so transform the function argument accordingly to change the integral over [1,5] to an equivalent one over [-1,1].

$x=2t+3\iff t=\dfrac x2-\dfrac32\implies2\mathrm dt=\mathrm dx$
$x=1\implies t=\dfrac{2-6}4=-1$
$x=5\implies t=\dfrac{10-6}4=1$

So,

$\displaystyle\int_{x=1}^{x=5}\sin x^2\,\mathrm dx=\displaystyle2\int_{t=-1}^{t=1}\sin(2t+3)^2\,\mathrm dt$

Let $f(t)=2\sin(2t+3)^2$ . With $n=4$ , we're looking for coefficients $c_i$ and nodes $x_i$ , with $1\le i\le4$ , such that

$\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx c_1f(x_1)+\cdots+c_4f(x_4)$

You can either try solving for each with the help of a calculator, or look up the values of the weights and nodes (they're extensively tabulated, and I'll include a link to one such reference).

Using the quadrature, we then have

$\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx0.3749f(-0.8611)+0.6521f(-0.3400)+0.6521f(0.3400)+0.3749f-0.8611)$
$\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx0.5790$

4 0

3 years ago