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Scorpion4ik [409]

3 years ago

10

(-6)3= ?can I get the answer plz

1 answer:

BARSIC [14]3 years ago

7 0

Answer:

-18 is the answer

Step-by-step explanation:

Rules in Multiplication and Division of Integers.

RULE 1: The product of a positive integer and a negative integer is negative.

RULE 2: The product of two positive integers is positive.

RULE 3: The product of two negative integers is positive.

So if you multiply 6 and 3 it equals 18 but there is an opposite sign

(+) (+) = (+)

(-) (-) = (+)

(+) (-) = (-)

(-) (+) = (-)

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Suppose KLMN ~ PQRS. Find the value of x.

arlik [135]

Complete question is attached.

Answer:

B) 8

Step-by-step explanation:

In this question, KLMN ~ PQRS. This means they are similar shapes.

From the image, we could denote that:

KL ~ PQ

LM ~ QR

Given that:

KL = 12

PQ = 30

QR = 20

LM = x

To find x, let's use tge expression:

$\frac{KL}{PQ} = \frac{LM}{QR}$

$= \frac{12}{30} = \frac{x}{20}$

Cross multiplying, we have:

12 * 20 = 30x

240 = 30x

To find x, let's divide both sides by 30.

$\frac{240}{30} = \frac{30x}{30}$

8 = x

x = 8

The correct option is option B.

3 0

3 years ago

What is the boundedness

DedPeter [7]

Answer:

Boundedness requires that there is only a bounded number of different kinds of parts

Step-by-step explanation:

8 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

Solve the following equation for x, in terms of a and b <br><br> ax= 15+bx

ZanzabumX [31]

Ax=15+bx
ax-bx=15
x(a-b)=15
x=15/(a-b)

4 0

3 years ago

A jar contains 18 gold coins, 36 rose gold coins, 9 silver coins, and 27 copper coins. What is the probability of randomly selec

nignag [31]

The answer 63% I worked it out

5 0

2 years ago