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2 years ago

Square contains 4cm. find the area of the shape.

Mathematics

2 answers:

mart [117]2 years ago

8 0

If the length is 4cm, it's probably 16. But just to check can you describe the question more?

Leya [2.2K]2 years ago

7 0

Answer:

Step-by-step explanation:

'Area of a square' formula is A = s^2, where s in the length of one side.

Here, with s = 4 cm, A = 16 cm^2

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Chloe and tino have a combined age of 48. three years ago chloe was double the age tino is now. find tinos age

jeka94

As given Chole and Tinos age , the present age of Tino is equal to 15 years.

As given,

Let x and y be the present age of Chole and Tino. respectively.

Combined age = 48

Given condition,

x + y = 48 _______(1)

Three years ago

x - 3 = 2y

⇒x =2y +3 _______(2)

Substitute the value of x in (1),

2y+3 +y =48

⇒ 3y = 45

⇒ y = 15years

Therefore , as given Chole and Tinos age , the present age of Tino is equal to 15 years.

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1 year ago

What does “4(x+3)+5x-x^2 “equal

Nikitich [7]

Answer:

Thanks for asking!

Step-by-step explanation:

The answer is -x²+9+12

5 0

2 years ago

Find the coordinates of the midpoints of the segment whose endpoints are H(4,8) and K (10,10)

creativ13 [48]

Answer:

(7,9)

Step-by-step explanation:

Use the midpoint formula to find the midpoint of the line segment.

4 0

2 years ago

Given 5n² + 6n + 7 = n² - 4n : a. Find the value of the discriminant. b. State the number and type of roots/solutions/zeros.[2 r

Vika [28.1K]

$\bf 5n^2+6n+7=n^2-4n\implies 5n^2+6n+7-n^2+4n=0
\\\\\\
4n^2+10n+7=0\\\\
-------------------------------\\\\
\begin{array}{llccll}
&{{ 4}}x^2&{{ +10}}x&{{ +7}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}
\\\\\\
discriminant\implies b^2-4ac=
\begin{cases}
0&\textit{one solution}\\
positive&\textit{two solutions}\\
negative&\textit{no solution}
\end{cases}$

if you get a positive value, that means 2 real solutions
if you get 0, means 1 real solution only
if you get a negative value, is an imaginary root solution, same as saying no solution.

4 0

2 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

Square contains 4cm. find the area of the shape.​

Square contains 4cm. find the area of the shape.