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3 years ago

6

(will give brainliest)+which is the same as moving a decimal point 3 places to the left in a decimal number

2 answers:

abruzzese [7]3 years ago

5 0

Hello!

I Believe multiplying by 1,000 is the same as moving a decimal point 3 places.

i hope it helps!

noname [10]3 years ago

4 0

12345.0 = 12.345 because you are moving the decimal place to the left 3 places

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Find all numbers such that three times the number is greater than eight less than the number.

klio [65]

Answer:

ALL SUCH NUMBERS WHICH ARE GREATER THAN -4 satisfy the given condition.

Step-by-step explanation:

Here, assume such number = m

Now, given :

3 times the number = 3 x ( m) = 3 m

8 less than the given number = m - 8

Now, according to the question:

3 times the number > 8 less than the given number

or, 3 m > m - 8

or, 3m - m > m - 8 + m

or, 2 m > - 8

or, m > - 4

Hence, all SUCH NUMBERS WHICH ARE GREATER THAN -4 satisfy the given condition.

6 0

3 years ago

A kintted pot holder is shaped like a circle. It's radius is 4.5 in. What is the area?

lesya [120]

Answer:

A =63.585 in^2

Step-by-step explanation:

The area of a circle is found by

A = pi r^2

A = pi (4.5)^2

A = pi(20.25)

We can approximate pi by 3.14

A = (3.14) (20.25)

A =63.585 in^2

8 0

3 years ago

Read 2 more answers

Marta is making 3 servings of fruit salad.she adds 3\8 cup blueberries for each serving. Her measuring cup holds 1\8 cup. How ma

Vilka [71]

She will get 3 because if you multiply 1/8 and 3 you will get an answer of 3/8

4 0

3 years ago

Please help meeeee! <br> I'm really in a rush rn!

yawa3891 [41]

Answer:

-3.392

Step-by-step explanation:

2.34

1.052

+_____

= 3.392

3 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