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3 years ago

5

A dog weighs two pounds less than three times the weight of a cat. The dog also weights twenty-two more pounds than the cat. Wri

te and solve an equation to find the weights of the cat and the dog.

1 answer:

Solnce55 [7]3 years ago

4 0

Answer:

the cat weighs 21 pounds.

the dog weighs 43 pounds

Step-by-step explanation:

dog = 3x-20 = x + 22

cat = x

3x-20 = x+22

3x = x+42 (add 20 to both sides)

2x = 42 (subtract x from both sides)

x = 21

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If T(x, y) = (x + 5, y + 6) and Pris the image of P, what is the rule for the translation in which P is the image of P'? T(x, y)

xxTIMURxx [149]

Answer:

The translation is;

$T(x,y)=(x-5,y-6)$

Explanation:

Given the translation rule;

$T(x,y)=(x+5,y+6)$

when P' is the image of P.

For the inverse, when P is the image of P', the Translation rule would become;

$\begin{gathered} x=x^{\prime}+5 \\ x^{\prime}=x-5 \\ y=y^{\prime}+6 \\ y^{\prime}=y-6 \\ So,\text{ the translation rule becomes;} \\ T(x,y)=(x-5,y-6) \end{gathered}$

The translation is;

$T(x,y)=(x-5,y-6)$

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1 year ago

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Step-by-step explanation:

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2 years ago

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3 years ago

Use the method of reduction of order to find a second solution to t^2y' + 3ty' – 3y = 0, t> 0 Given yı(t) = t y2(t) = Preview

Anestetic [448]

Let $y_2(t)=tv(t)$ . Then

${y_2}'=tv'+v$

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and substituting these into the ODE gives

$t^2(tv''+2v')+3t(tv'+v)-3tv=0$

$t^3v''+5t^2v'=0$

$tv''+5v'=0$

Let $u(t)=v'(t)$ , so that $u'(t)=v''(t)$ . Then the ODE is linear in $u$ , with

$tu'+5u=0$

Multiply both sides by $t^4$ , so that the left side can be condensed as the derivative of a product:

$t^5u'+5t^4u=(t^5u)'=0$

Integrating both sides and solving for $u(t)$ gives

$t^5u=C\implies u=Ct^{-5}$

Integrate again to solve for $v(t)$ :

$v=C_1t^{-6}+C_2$

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3 years ago

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valkas [14]

Answer:

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Step-by-step explanation:

Given the bean bag made by beth is in the shape of cube with each side of the cube equal to 2 feet.

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Now the volume of the bean bag made by beth is

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3 years ago