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3 years ago

Composite number between 70 and 80

Mathematics

2 answers:

NeX [460]3 years ago

4 0

The composite numbers are 72, 74,77 and 78

Sever21 [200]3 years ago

4 0

The answer's are 72, 74, 77, and 78.

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The diameter and height of this cylinder are equal to the side length, S, of the cube in which the cylinder is inscribed. What i

sammy [17]

Volume = πr²h

Diameter = S

Height = S

Radius = S/2

Volume = π ( s²/4 ) s

1/4 πs³

The required answer is in the option ( A )

Hope helps!

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

PQR is an isosceles triangle in which PQ = PR

otez555 [7]

Answer and Step-by-step explanation: Congruent triangles are triangles with the same three sides and same three angles.

There many ways to determine if 2 triangles are congruent.

One of them is ASA or Angle, Side, Angle and it means that if two angles and the included side of one triangle are equal to the corresponding angles and side on the other triangle, they are congruent.

In this case, angle MRQ and angle NQR are equal. The included side of both triangles are the same QR, so it can be concluded that triangle QNR is congruent to triangle RMQ.

The image in the attachment shows the angles and their included side, which are colored.

3 0

3 years ago

Write two expressions that could be used to find 105% of x. ANSWER TODAY!! :) thanks

Margaret [11]

First way: x * 1.05

This computes directly what 105% of x will be.

Second way: x + (x * 0.05)

This takes x (which is 100%) and adds 5% of x to it.

6 0

3 years ago

Read 2 more answers

If M is the midpoint of XY, find the coordinates of X if M(-3, -1) and Y(-8, 6) A. (2.5, -5.5) B. (-5.5, 2.5) C. (0, 0) D. (2, -

nadezda [96]

Answer:

(2, -8)

Step-by-step explanation:

(x, y)=(-3, -1)

(x1+x2)/2=-3,

(y2+y2)/2=-1,

-------------------

(x1+x2)/2=-3

(y1+y2)/2=-1

---------------------

x1-8=-6

x1=2

----------

y1+6=-2

y1=-8

(2, -8)

6 0

2 years ago