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3 years ago

Help me please help ASAP help

Mathematics

1 answer:

Tomtit [17]3 years ago

4 0

Answer:

98.1 unit²

Step-by-step explanation:

area of semi-circle= ½πr²

½π(3)²

= 9/2π

area of isosceles trapezoid= ½(b1+b2)h

½(15+6)8

=84

area of figure= 9/2π +84

98.1 unit²

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ASAP

Lerok [7]

Answer:$90,000

Step-by-step explanation:

3,000*30=90,000

4 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

-2{-7-2[3 squared + 2 to the power of 4]-8÷4}

Leto [7]

Answer:

118 i think

Step-by-step explanation:

-2{-7-2[3 squared + 2 to the power of 4]-8÷4}

-2{-7-2{9+16]-2}

-2{-7-50-2}

-2(-59)

118

5 0

3 years ago

If it costs $90 to feed a family of 3 for one week. (3 points) Unit 8 lesson 4

liubo4ka [24]

X = number of weeks
1 week =
3x = $90
Divide by 3
3x/3 = x
$90/3 = $30
1 person = $30 per week
x 5
5 people = $150 per week

x6
6 people = $180 per week

3 0

3 years ago

Find the greatest number that will divide 63, 45 and 69 so as to leave the same remainder.

jeka57 [31]

Answer:

Step-by-step explanation:

Let n be unknown divisor and a be the same remainder, then

$63=q_1\cdot n+a\\ \\45=q_2\cdot n+a\\ \\69=q_3\cdot n+a$