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2 years ago

Find the length of the diagonal of the rectangle.

Mathematics

1 answer:

Anuta_ua [19.1K]2 years ago

7 0

Answer:

≈ 11.4 m

Step-by-step explanation:

The diagonal splits the rectangle into 2 right triangles , with legs 7 and 9 and the diagonal the hypotenuse.

Using Pythagoras' identity with h as the diagonal, then

h² = 7² + 9² = 49 + 81 = 130 ( take the square root of both sides )

h = $\sqrt{130}$ ≈ 11.4 m ( to the nearest tenth )

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Rearrange the equation so r is the independent variable.<br><br> 10q - 5r = 30

Murljashka [212]

Answer:

q = 3 + 1/2r

Step-by-step explanation:

10q - 5r = 30

We want to solve for q

Add 5r to each side

10q - 5r+5r = 30+5r

10q = 30+5r

Divide by 10

10q/10 = 30/10 +5r/10

q = 3 + 1/2r

3 0

3 years ago

The surface of each crystal has an area of

Umnica [9.8K]

Answer:

10,000,000,000 × 10^-1, 1/1,000,000,000^-1

Step-by-step explanation:

Im not sure what you are asking

7 0

3 years ago

Calculus 3 help please.

Reptile [31]

I assume each path $C$ is oriented positively/counterclockwise.

(a) Parameterize $C$ by

$\begin{cases} x(t) = 4\cos(t) \\ y(t) = 4\sin(t)\end{cases} \implies \begin{cases} x'(t) = -4\sin(t) \\ y'(t) = 4\cos(t) \end{cases}$

with $-\frac\pi2\le t\le\frac\pi2$ . Then the line element is

$ds = \sqrt{x'(t)^2 + y'(t)^2} \, dt = \sqrt{16(\sin^2(t)+\cos^2(t))} \, dt = 4\,dt$

and the integral reduces to

$\displaystyle \int_C xy^4 \, ds = \int_{-\pi/2}^{\pi/2} (4\cos(t)) (4\sin(t))^4 (4\,dt) = 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt$

The integrand is symmetric about $t=0$ , so

$\displaystyle 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \,dt$

Substitute $u=\sin(t)$ and $du=\cos(t)\,dt$ . Then we get

$\displaystyle 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^1 u^4 \, du = \frac{2^{13}}5 (1^5 - 0^5) = \boxed{\frac{8192}5}$

(b) Parameterize $C$ by

$\begin{cases} x(t) = 2(1-t) + 5t = 3t - 2 \\ y(t) = 0(1-t) + 4t = 4t \end{cases} \implies \begin{cases} x'(t) = 3 \\ y'(t) = 4 \end{cases}$

with $0\le t\le1$ . Then

$ds = \sqrt{3^2+4^2} \, dt = 5\,dt$

and

$\displaystyle \int_C x e^y \, ds = \int_0^1 (3t-2) e^{4t} (5\,dt) = 5 \int_0^1 (3t - 2) e^{4t} \, dt$

Integrate by parts with

$u = 3t-2 \implies du = 3\,dt \\\\ dv = e^{4t} \, dt \implies v = \frac14 e^{4t}$

$\displaystyle \int u\,dv = uv - \int v\,du$

$\implies \displaystyle 5 \int_0^1 (3t-2) e^{4t} \,dt = \frac54 (3t-2) e^{4t} \bigg|_{t=0}^{t=1} - \frac{15}4 \int_0^1 e^{4t} \,dt \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} e^{4t} \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} (e^4 - 1) = \boxed{\frac{5e^4 + 55}{16}}$

$\begin{cases} x(t) = 3(1-t)+t = -2t+3 \\ y(t) = (1-t)+2t = t+1 \\ z(t) = 2(1-t)+5t = 3t+2 \end{cases} \implies \begin{cases} x'(t) = -2 \\ y'(t) = 1 \\ z'(t) = 3 \end{cases}$

with $0\le t\le1$ . Then

$ds = \sqrt{(-2)^2 + 1^2 + 3^2} \, dt = \sqrt{14} \, dt$

and

$\displaystyle \int_C y^2 z \, ds = \int_0^1 (t+1)^2 (3t+2) \left(\sqrt{14}\,ds\right) \\\\ ~~~~~~~~ = \sqrt{14} \int_0^1 \left(3t^3 + 8t^2 + 7t + 2\right) \, dt \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 t^4 + \frac83 t^3 + \frac72 t^2 + 2t\right) \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 + \frac83 + \frac72 + 2\right) = \boxed{\frac{107\sqrt{14}}{12}}$

8 0

1 year ago

1/2 to the power of 5 equals

svp [43]

Answer:

1/32

Step-by-step explanation:

This could be written as $1/2^5\\$

1/2 x 1/2 x 1/2 x 1/2 x 1/2

= 1/32, or 0.03125.

3 0

2 years ago

charity is four times as old as her daughter. she is also six years younger than her husband. The sum of their ages ten years to

Pachacha [2.7K]

Answer:

36years

Step-by-step explanation:

Let charity present age be x

Charity daughter present age be y

Charity husband present age be z

If the sum of their ages ten years to come is 117, then;

10+x+10+y+10+z = 117

30+x+y+z = 117

x+y+z = 87 ... 1

If charity is four times as old as her daughter, then;

x = 4y

y = x/4 ... 2

If she is also six years younger than her husband, then;

x = z- 6

z = x+6 .. 3

Substitute 2 and 3 into 1;

x + x/4 + (x+6) = 87

Multiply through by 4

4x + x + 4(x+6) = 4(87)

5x+4x+24 = 348

9x = 348 - 24

9x = 324

x = 324/9

x = 36

hence Charity is 36years old today

5 0

2 years ago