Can anyone help me with this please <br> Don’t answer if you do not know .

(b) The area of a rectangular painting is 7719 cm2.

Nataliya [291]

Given:

The area of a rectangular painting is 7719 cm²
Width of painting = 83cm

To Find?

Length of painting.

Solution:

Let length of painting be x cm

Using formula:

Area of rectangle = L × B

Where,

Length = x cm
Breadth = 73 cm

→ 7719 = x × 83

→ x = 7719/83

→ x = 93 cm

Hence,

Length of painting is 93cm

6 0

2 years ago

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1. James wants to fill a tank with water. He already has 4 gallons in the 50 gallon tank. Using the equation y=2x+4, where y is

rusak2 [61]

Y=2(4)+4
=12 gallons in the tank so far

50-12
=38 gallons to fill

3 0

3 years ago

Out of the 150 students at a summer camp, 72 signed up for canoeing. there were 23 students who signed up for trekking, and 13 o

True [87]

Answer:

Approximately 45% students signed up for neither canoeing or trekking.

Step-by-step explanation:

First we subtract the number of the students who signed up from the number of the total students to find the number of the students who did not signed up for either.

But there are 13 students who signed up canoeing and also for trekking.

Hence the number of students they signed up for activity is

72 + 23 - 13 = 82

And the number of students they not signed up for any activity is

150 - 82 = 68

so 68 students signed up neither for trekking nor canoeing. The percentage of those students are :

$\frac{68}{150}$ × 100 = 45.3333%

Approximately 45% students signed up for neither canoeing or trekking.

8 0

3 years ago

Find a particular solution to <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7B%20d%5E%7B2%7Dy%20%7D%7Bd%20x%5E%7

Digiron [165]

$y=x^r$
$\implies r(r-1)x^r+6rx^r+4x^r=0$
$\implies r^2+5r+4=(r+1)(r+4)=0$
$\implies r=-1,r=-4$

so the characteristic solution is

$y_c=\dfrac{C_1}x+\dfrac{C_2}{x^4}$

As a guess for the particular solution, let's back up a bit. The reason the choice of $y=x^r$ works for the characteristic solution is that, in the background, we're employing the substitution $t=\ln x$ , so that $y(x)$ is getting replaced with a new function $z(t)$ . Differentiating yields

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dz}{\mathrm dt}$
$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac1{x^2}\left(\dfrac{\mathrm d^2z}{\mathrm dt^2}-\dfrac{\mathrm dz}{\mathrm dt}\right)$

Now the ODE in terms of $t$ is linear with constant coefficients, since the coefficients $x^2$ and $x$ will cancel, resulting in the ODE

$\dfrac{\mathrm d^2z}{\mathrm dt^2}+5\dfrac{\mathrm dz}{\mathrm dt}+4z=e^{2t}\sin e^t$

Of coursesin, the characteristic equation will be $r^2+6r+4=0$ , which leads to solutions $C_1e^{-t}+C_2e^{-4t}=C_1x^{-1}+C_2x^{-4}$ , as before.

Now that we have two linearly independent solutions, we can easily find more via variation of parameters. If $z_1,z_2$ are the solutions to the characteristic equation of the ODE in terms of $z$ , then we can find another of the form $z_p=u_1z_1+u_2z_2$ where

$u_1=-\displaystyle\int\frac{z_2e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$
$u_2=\displaystyle\int\frac{z_1e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$

where $W(z_1,z_2)$ is the Wronskian of the two characteristic solutions. We have

$u_1=-\displaystyle\int\frac{e^{-2t}\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_1=\dfrac23(1-2e^{2t})\cos e^t+\dfrac23e^t\sin e^t$

$u_2=\displaystyle\int\frac{e^t\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_2=\dfrac13(120-20e^{2t}+e^{4t})e^t\cos e^t-\dfrac13(120-60e^{2t}+5e^{4t})\sin e^t$

$\implies z_p=u_1z_1+u_2z_2$
$\implies z_p=(40e^{-4t}-6)e^{-t}\cos e^t-(1-20e^{-2t}+40e^{-4t})\sin e^t$

and recalling that $t=\ln x\iff e^t=x$ , we have

$\implies y_p=\left(\dfrac{40}{x^3}-\dfrac6x\right)\cos x-\left(1-\dfrac{20}{x^2}+\dfrac{40}{x^4}\right)\sin x$

4 0

2 years ago

The mean of 20 items is 40 and the mean of

kipiarov [429]

Answer:

At this point just choose a mean thats half of the value

5 0

3 years ago

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Can anyone help me with this please Don’t answer if you do not know .