All categories

3 years ago

9. Given that P = (-1,16) and Q = (1, 9), find the component form and

1 answer:

7 0

To find the component form, u only need to know how to substitute figures for letters. What do I mean by this?

If initial side is (x1,y1)
Then x1 = -1, and y 1 = 16
If the terminal side is x2, y2
Then x2 =1 and y2 =9

This component from p is x2-x1 y2-y1
In this case <[1-(-1)], (9-16)
Which give us
2,-7

If p = x ,y using Pythagorean’s theorem
2(2) + (-7)(7)
4+49
53
Square root of 53 is
7.28010988928
7.3

You might be interested in

ULLIVLY/1/2/3/assessment

zvonat [6]

<h3>The worth after 4 years is $ 680.24</h3>

<em><u>Solution:</u></em>

<em><u>The formula for compound interest, including principal sum, is:</u></em>

$A = p(1+\frac{r}{n})^{nt}$

Where,

A = the future value of the investment

P = the principal investment amount

r = the annual interest rate (decimal)

n = the number of times that interest is compounded per unit t

t = the time the money is invested

From given,

n = 1 ( since interest is compounded annually)

p = 500

t = 4

$r = 8 \% = \frac{8}{100} = 0.08$

<em><u>Substituting the values we get,</u></em>

$A = 500(1+ \frac{0.08}{1})^{1 \times 4}\\\\A = 500(1.08)^4\\\\A = 500 \times 1.36048896\\\\A = 680.24448 \approx 680.24$

Thus the worth after 4 years is $ 680.24

3 0

3 years ago

How do you do this question?

Lena [83]

Step-by-step explanation:

The Taylor series expansion is:

Tₙ(x) = ∑ f⁽ⁿ⁾(a) (x − a)ⁿ / n!

f(x) = 1/x, a = 4, and n = 3.

First, find the derivatives.

f⁽⁰⁾(4) = 1/4

f⁽¹⁾(4) = -1/(4)² = -1/16

f⁽²⁾(4) = 2/(4)³ = 1/32

f⁽³⁾(4) = -6/(4)⁴ = -3/128

Therefore:

T₃(x) = 1/4 (x − 4)⁰ / 0! − 1/16 (x − 4)¹ / 1! + 1/32 (x − 4)² / 2! − 3/128 (x − 4)³ / 3!

T₃(x) = 1/4 − 1/16 (x − 4) + 1/64 (x − 4)² − 1/256 (x − 4)³

f(x) = 1/x has a vertical asymptote at x=0 and a horizontal asymptote at y=0. So we can eliminate the top left option. That leaves the other three options, where f(x) is the blue line.

Now we have to determine which green line is T₃(x). The simplest way is to notice that f(x) and T₃(x) intersect at x=4 (which makes sense, since T₃(x) is the Taylor series centered at x=4).

The bottom right graph is the only correct option.

3 0

3 years ago

"Suppose an object falling in the atmosphere has mass m=15kg and the drag coefficient is γ=9kg/s. Recall that the differential e

Art [367]

Answer:

a. v(t)= -6.78 $e^{-16.33t}$ + 16.33 b. 16.33 m/s

Step-by-step explanation:

The differential equation for the motion is given by mv' = mg - γv. We re-write as mv' + γv = mg ⇒ v' + γv/m = g. ⇒ v' + kv = g. where k = γ/m.Since this is a linear first order differential equation, We find the integrating factor μ(t)= $e^{\int\limits^ {}k \, dt }$ = $e^{kt}$ . We now multiply both sides of the equation by the integrating factor.

μv' + μkv = μg ⇒ $e^{kt}$ v' + k $e^{kt}$ v = g $e^{kt}$ ⇒ [v $e^{kt}$ ]' = g $e^{kt}$ . Integrating, we have

∫ [v $e^{kt}$ ]' = ∫g $e^{kt}$

v $e^{kt}$ = $\frac{g}{k}$ $e^{kt}$ + c

v(t)= $\frac{g}{k}$ + c $e^{-kt}$ .

From our initial conditions, v(0) = 9.55 m/s, t = 0 , g = 9.8 m/s², γ = 9 kg/s , m = 15 kg. k = y/m. Substituting these values, we have

9.55 = 9.8 × 15/9 + c $e^{-16.33 * 0}$ = 16.33 + c

c = 9.55 -16.33 = -6.78.

So, v(t)= 16.33 - 6.78 $e^{-16.33t}$ . m/s = - 6.78 $e^{-16.33t}$ + 16.33 m/s

b. Velocity of object at time t = 0.5

At t = 0.5, v = - 6.78 $e^{-16.33 x 0.5}$ + 16.33 m/s = 16.328 m/s ≅ 16.33 m/s

6 0

3 years ago

3 regions are defined in the figure find the volume generated by rotating the given region about the specific line

anastassius [24]

The volume generated by rotating the given region $R_{3}$ about OC is $\frac{4}{g} \pi$

<h3>Washer method</h3>

Because the given region ( $R_{3}$ ) has a look like a washer, we will apply the washer method to find the volume generated by rotating the given region about the specific line.

solution

We first find the value of x and y

$y=2(x)^{\frac{1}{4} }$

$x=(\frac{y}{2} )^{4}$

$y=2x$

$x=\frac{y}{2}$

$\int\limits^a_b {\pi } \, (R_{o^{2} } - R_{i^{2} } ) dy$

$R_{o} = x = \frac{y}{2}$

$R_{i} = x= (\frac{y}{2}) ^{4}$

$a=0, b=2$

$v= \int\limits^2_o {\pi } \, [(\frac{y}{2})^{2} - ((\frac{y}{2}) ^{4} )^{2} ) dy$

$v= \pi \int\limits^2_o= [\frac{y^{2} }{4} - \frac{y^{8} }{2^{8} }} ] dy$

$v= \pi [\int\limits^2_o {\frac{y^{2} }{4} } \, dy - \int\limits^2_o {\frac{y}{2^{8} } ^{8} } \, dy ]$

$v=\pi [\frac{1}{4} \frac{y^{3} }{3} \int\limits^2_0 - \frac{1}{2^{8} } \frac{y^{g} }{g} \int\limits^2_o\\v= \pi [\frac{1}{12} (2^{3} -0)-\frac{1}{2^{8}*9 } (2^{g} -0)]\\v= \pi [\frac{2}{3} -\frac{2}{g} ]\\v= \frac{4}{g} \pi$

A similar question about finding the volume generated by a given region is answered here: brainly.com/question/3455095

6 0

2 years ago

X/5 + 3 = 9 <br> solve equation

Delicious77 [7]

Answer:

X= 15

Step-by-step explanation:

multiply 5 on x/5 and 9

divide 3 and 45

x=15

3 0

3 years ago