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3 years ago

5

Help please ( ˶ ❛ ꁞ ❛ ˶ )

2 answers:

pentagon [3]3 years ago

6 0

Answer:

179.25 cm squared

Step-by-step explanation:

14*10=140

3.14(5)^2=78.5/2=39.25

140+39.25=179.25

VashaNatasha [74]3 years ago

4 0

179.25in squared is the answer (press thanks plz)

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-950-?= -379<br><br><br> Help Will give brainliest!!

Karolina [17]

Answer: -571

Step-by-step-explanation:
-950 - (-571) = -379
-950 + 571 = -379
Hope this helps :)

8 0

2 years ago

What is the sum of an infinite geometric series if the first term is 81 and the common ratio is 2⁄3?

Marrrta [24]

C

the sum to infinity of a geometric series = $\frac{a}{1-r}$

where a is the first term and r the common ratio

= $\frac{81}{\frac{1}{3} }$ = 81 × 3 = 243

4 0

3 years ago

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

Find the ratio of the following 1) 25to 35

Ad libitum [116K]

Answer:

The ratio of 25/35 is

Step-by-step explanation:

25/35

=5/7 (both cut by 5)

So, your answer is 5:7

8 0

3 years ago

Read 2 more answers

The graph of g(x)=(0.5)x+4 is shown. Which equation is an asymptote of this function? y = 0 y = 4 x = 4 x = 0

Alexandra [31]

we are given

$g(x)=(0.5)^x+4$

we can see that

there is no value of x for which g(x) is not defined

so, no vertical asymptote exists

now, we will find horizontal asymptote

$\lim_{x \to \infty} g(x)= \lim_{x \to \infty}((0.5)^x+4 )$

$\lim_{x \to \infty} g(x)= (\lim_{x \to \infty} (0.5)^x+\lim_{x \to \infty} 4 )$

$\lim_{x \to \infty} g(x)= 0+4$

so, we get

horizontal asymptote as

$y= 4$ ............Answer

3 0

2 years ago