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2 years ago

Please help i need now

Mathematics

2 answers:

Sunny_sXe [5.5K]2 years ago

7 0

Answer:

8x10^-12

Step-by-step explanation:

Hope that helps.

laila [671]2 years ago

6 0

Answer:

the answer is 8*10^-12

Step-by-step explanation:

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Please find x and y <br><br>x=y-4 and -2x+3y=6

laiz [17]

Answer:

x=y-4 is x=y-4

y=2+ 2x/2

Step-by-step explanation:

3 0

2 years ago

A different map has a scale of 1:50000

Deffense [45]

Answer:

8 Km

Step-by-step explanation:

16 x 50000 = 800,000 cm

800,000 cm = 8,000 m

8,000 m = 8 km

I hope this helps, if not sorry

6 0

2 years ago

Find the inverse of the function f(x) = (x - 4) 2 - 5 if x ≥ 4.

Serggg [28]

Same here, we do a quick switcharoo on the variables first,

$\bf \stackrel{f(x)}{y}=(x-4)^2-5\qquad inverse\implies \boxed{x}=\left( \boxed{y}-4 \right)^2-5
\\\\\\
x+5=(y-4)^2\implies \pm\sqrt{x+5}=y-4\implies \pm\sqrt{x+5}+4=y$

6 0

2 years ago

A coffee shop covers it coffee cups for holiday seasons. Their cups are cylindrical, have a 2 inch radius, and a height of 6 inc

slega [8]

Answer:

Surface area of square box cover is 4x12=48 $in^{2}$

Step-by-step explanation:

Given cylindrical cup have 2 in of radius and height of 6 in.

It is said that we need to cover the cup leaving top and bottom of cup.

To find how much material to cover:

After covering the cup, square box will be formed around the cup.

As shown in figure, Top view of square cup and box.

One face of box has length of diameter of cup = 2 in and height of box as height of cup = 6 in

Therefore, Area of one face of square box is length x height = 2 x 6 =12 $in^{2}$

Since, Box having 4 faces

Surface area of square box cover = 4 x area of one face of box

= 4x12=48 $in^{2}$

8 0

3 years ago

Consider the differential equation:

Wewaii [24]

(a) Take the Laplace transform of both sides:

$2y''(t)+ty'(t)-2y(t)=14$

$\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s$

where the transform of $ty'(t)$ comes from

$L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)$

This yields the linear ODE,

$-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s$

Divides both sides by $-s$ :

$Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}$

Find the integrating factor:

$\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C$

Multiply both sides of the ODE by $e^{3\ln|s|-s^2}=s^3e^{-s^2}$ :

$s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}$

The left side condenses into the derivative of a product:

$\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}$

Integrate both sides and solve for $Y(s)$ :

$s^3e^{-s^2}Y(s)=7e^{-s^2}+C$

$Y(s)=\dfrac{7+Ce^{s^2}}{s^3}$

(b) Taking the inverse transform of both sides gives

$y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]$

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that $\frac{7t^2}2$ is one solution to the original ODE.

$y(t)=\dfrac{7t^2}2\implies y'(t)=7t\implies y''(t)=7$

Substitute these into the ODE to see everything checks out:

$2\cdot7+t\cdot7t-2\cdot\dfrac{7t^2}2=14$

5 0

3 years ago