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3 years ago

7

How much more is five ninth’s greater than 4 fifteenths

1 answer:

andrey2020 [161]3 years ago

8 0

The answer is 3.46154

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GOOD AT MATH!! PLEASE HELPPP

Valentin [98]

It’s B I’m pretty sure

8 0

3 years ago

Read 2 more answers

The ratio of the width to the length of a rectangle is 2:3, respectively. Answer each of the following. a By what percent would

Elina [12.6K]

Answer:

The percentage increase in Area of rectangle is 200%

Step-by-step explanation:

Given as :

The ratio of the width to the length of a rectangle is 2:3

Let The length of rectangle = 2 x

Let The width of rectangle = 3 x

∵ Area of rectangle = length × width

So, $A_1$ = 2 x × 3 x

Or, $A_1$ = 6 x²

<u>Again</u>

The increased length of rectangle = 2 x + 2 x = 4 x

The increase width of rectangle = 3 x + 50% of 3 x

I.e The increase width of rectangle = 3 x + 1.5 x = 4.5 x

∵ Increased Area of rectangle = increased length × increased width

Or, $A_2$ = 4 x × 4.5 x = 18 x²

So, The percentage increase in area = $\dfrac{A_2 - A_1}{A_1}$ × 100

Or, The percentage increase in area = $\frac{18 x^{2}-6x^{2}}{6x^{2}}$ × 100

Or , The percentage increase in area = $\dfrac{12}{6}$ × 100

∴ The percentage increase in area = 200

Hence, The percentage increase in Area of rectangle is 200% Answer

5 0

3 years ago

Which product is positive?<br> BA :)<br> OGO<br> (100)<br> G<br> 8<br> 3

romanna [79]

Wouldn’t the answer be 8 and 3

6 0

3 years ago

Simplify 2(5x – 4) + 3x

guapka [62]

To simplify, you must distribute the 2 into the 5x and -4 by multiplying 2 with 5x and 2 with -4.

<span>2(5x – 4) + 3x

2 x 5x = 10x

2 x -4 = -8

You will get 10x - 8 + 3x

Now you must combine like terms or add like terms. In this case we must add 10x and 3x

10x + 3x = 13x

Finally the answer would be 13x - 8</span>

6 0

3 years ago

Read 2 more answers

Apply the method of undetermined coefficients to find a particular solution to the following system.wing system.

jarptica [38.1K]

$y''-y'+y=\sin x$

The corresponding homogeneous ODE has characteristic equation $r^2-r+1=0$ with roots at $r=\dfrac{1\pm\sqrt3}2$ , thus admitting the characteristic solution

$y_c=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x$

For the particular solution, assume one of the form

$y_p=a\sin x+b\cos x$

${y_p}'=a\cos x-b\sin x$

${y_p}''=-a\sin x-b\cos x$

Substituting into the ODE gives

$(-a\sin x-b\cos x)-(a\cos x-b\sin x)+(a\sin x+b\cos x)=\sin x$

$-b\cos x+a\sin x=\sin x$

$\implies a=1,b=0$

Then the general solution to this ODE is

$\boxed{y(x)=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x+\sin x}$

$y''-3y'+2y=e^x\sin x$

$\implies r^2-3r+2=(r-1)(r-2)=0\implies r=1,r=2$

$\implies y_c=C_1e^x+C_2e^{2x}$

Assume a solution of the form

$y_p=e^x(a\sin x+b\cos x)$

${y_p}'=e^x((a+b)\cos x+(a-b)\sin x)$

${y_p}''=2e^x(a\cos x-b\sin x)$

Substituting into the ODE gives

$2e^x(a\cos x-b\sin x)-3e^x((a+b)\cos x+(a-b)\sin x)+2e^x(a\sin x+b\cos x)=e^x\sin x$

$-e^x((a+b)\cos x+(a-b)\sin x)=e^x\sin x$

$\implies\begin{cases}-a-b=0\\-a+b=1\end{cases}\implies a=-\dfrac12,b=\dfrac12$

so the solution is

$\boxed{y(x)=C_1e^x+C_2e^{2x}-\dfrac{e^x}2(\sin x-\cos x)}$

$y''+y=x\cos(2x)$

$r^2+1=0\implies r=\pm i$

$\implies y_c=C_1\cos x+C_2\sin x$

Assume a solution of the form

$y_p=(ax+b)\cos(2x)+(cx+d)\sin(2x)$

${y_p}''=-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x)$

Substituting into the ODE gives

$(-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x))+((ax+b)\cos(2x)+(cx+d)\sin(2x))=x\cos(2x)$

$-(3ax+3b-4c)\cos(2x)-(3cx+3d+4a)\sin(2x)=x\cos(2x)$

$\implies\begin{cases}-3a=1\\-3b+4c=0\\-3c=0\\-4a-3d=0\end{cases}\implies a=-\dfrac13,b=c=0,d=\dfrac49$

so the solution is

$\boxed{y(x)=C_1\cos x+C_2\sin x-\dfrac13x\cos(2x)+\dfrac49\sin(2x)}$

7 0

3 years ago