X - 3y = 0 and 3x + y = 7

Given c=(2.4,0.45) and d = (7,-4) find the direction of 5c+4d

kakasveta [241]

Answer:

The direction of 5c + 4d is approximately 19.0° clockwise with

the positive part of x-axis

Step-by-step explanation:

* Lets talk about the direction of a vector

- If the vector is (x , y), then

# Its magnitude is √(x² + y²)

# Its direction is tan^-1 (y/x)

- The direction is the angle between the positive part of x-axis and

the vector

* Lets solve the problem

∵ c = (2.4 , 0.45) and d = (7 , -4)

- To find 5c + 4d , multiply the two coordinates of c by 5 and the two

coordinates of d by 4

# Multiply c by 5

∴ 5c = [5(2.4) , 5(0.45)]

∴ 5c = (12 , 2.25)

# Multiply d by 4

∴ 4d = [4(7) , 4(-4)]

∴ 4d = (28 , -16)

- Lets add 5c and 4d

∴ 5c + 4d = (12 , 2.25) + (28 , -16)

∴ 5c + 4d = [(12 + 28) , (2.25 + -16)]

∴ 5c + 4d = (40 , -13.75)

- The vector is in the 4th quadrant because the x-coordinate is

positive and the y-coordinate is negative

* Find the direction of 5c + 4d

∵ The direction is tan^-1 (y/x)

∴ The direction is tan^-1 (-13.75/40) = -18.97°

* The direction of 5c + 4d is approximately 19.0° clockwise with

the positive part of x-axis

3 0

3 years ago

Megan was painting on a rectangular canvas, and needed a frame for it. The length of the frame is 15 inches and the width is 10

siniylev [52]

Answer:

50 inches

Step-by-step explanation:

Hi there!

$P=2l+2w$ where l is the length of the rectangle and w is the width of the rectangle

Plug in the length (15 inches) and width (10 inches)

$P=2(15)+2(10)\\P=30+20\\P=50$

Therefore, the perimeter of the frame is 50 inches.

I hope this helps!

8 0

2 years ago

EASY 50 POINTS. PLEASE HELP!!!

Y_Kistochka [10]

Answer:

y= (x+5)^2 +7

Step-by-step explanation:

4 0

3 years ago

PLEASE HELP ASAP!!

Alekssandra [29.7K]

I think its D. 32.6 s/ft and appoximatley 0.03 ft/s

8 0

3 years ago

Let f be defined by the function f(x) = 1/(x^2+9)

riadik2000 [5.3K]

(a)

$\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}$

Substitute x = 3 tan(t ) and dx = 3 sec²(t ) dt :

$\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt$

$=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}$

(b) The series

$\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}$

converges by comparison to the convergent p-series,

$\displaystyle\sum_{n=3}^\infty\frac1{n^2}$

(c) The series

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}$

converges absolutely, since

$\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}$

That is, ∑ (-1)ⁿ (n ² + 9)/eⁿ converges absolutely because ∑ |(-1)ⁿ (n ² + 9)/eⁿ| = ∑ (n ² + 9)/eⁿ in turn converges by comparison to a geometric series.

5 0

3 years ago