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2 years ago

Trình bày một số ứng dụng của công thức cộng , công thức nhân trong ghép nối các thiết bị để tăng độ tin cậy trong hệ thống

Mathematics

1 answer:

liraira [26]2 years ago

8 0

Answer:

ano po pakitagalog Po pls Hindi ko po kasi maintindihan

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LaShawn solved the equation below to the determine the solution.

alexira [117]

Answer:

x = all real numbers.

Step-by-step explanation:

3 x minus 8 = negative x + 4 (x minus 2)

3x - 8 = -x + 4(x - 2)

3x - 8 = -x + 4x - 8

3x - 8 = 3x - 8

3x - 3x = -8 + 8

0 = 0

Since the result is a true statement, but 0 = 0, x is equivalent to all real numbers.

Hope this helps!

5 0

2 years ago

Read 2 more answers

You start at (2, -4). You move up 4 units. Where do you end?

NeTakaya

Answer:

(2, 0)

Step-by-step explanation:

Hope This Helps! : )

7 0

2 years ago

NO LINKS OR FILES!

Archy [21]

(a) If the particle's position (measured with some unit) at time t is given by s(t), where

$s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}$

then the velocity at time t, v(t), is given by the derivative of s(t),

$v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}$

(b) The velocity after 3 seconds is

$v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}$

$\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}$

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

$\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|$

In interval notation, this happens for t in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt$

By definition of absolute value, we have

$|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)$

In part (d), we've shown that v(t) > 0 when -√11 < t < √11, so we split up the integral at t = √11 as

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt$

and by the fundamental theorem of calculus, since we know v(t) is the derivative of s(t), this reduces to

$s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}$

7 0

2 years ago

Factor 3x^3+6x^2+x+2 by grouping

bezimeni [28]

The answer for this question when grouping:

(x+2)(3x^2+1)

7 0

2 years ago

Read 2 more answers

Lori creates the following design for a T-shirt.

kakasveta [241]

Solution-

From the figure,

AE = 2.4

EB = 2.8

BC = 11.7

Area of rectangle 1 = 8.68 sq.in

$\Rightarrow FH \times HI=8.68$

$\Rightarrow EB \times HI=8.68$ (∵ sides of the rectangle 2)

$\Rightarrow 2.8 \times HI=8.68$

$\Rightarrow HI=3.1$

Area of Triangle 1 = 6.48 sq.in

$\Rightarrow \frac{1}{2}\times AE \times EG= 6.48$

$\Rightarrow \frac{1}{2}\times AE \times (EF+FG)= 6.48$

$\Rightarrow \frac{1}{2}\times AE \times (EF+HI)= 6.48$ (∵ sides of the rectangle 1)

$\Rightarrow EF+3.1= 5.4$

$\Rightarrow EF=2.3$

$\Rightarrow BH=2.3$ (∵ sides of the rectangle 2)

$BC = BH+HI+IC$

$\Rightarrow 11.7= 2.3+3.1+IC$

$\Rightarrow IC=6.3$

The area of Rectangle 2,

$=EB\times BH =2.8\times 2.3=6.44\ sq.in$

The area of Triangle 2,

$\frac{1}{2}\times GI \times IC=\frac{1}{2}\times EB \times IC=\frac{1}{2}\times 2.8 \times 6.3=8.82\ sq.in$

The area of the whole figure = Area of Triangle 1 + Area of rectangle 1 + Area of Triangle 2 + Area of rectangle 2

= 6.48+8.68+8.82+6.44=30.42 sq.in

6 0

2 years ago