Which of the following describes this?

Answer:

Yes, the expressions are equivalent.

Step-by-step explanation:

Expanding (3x + 9)(x + 2) gives us [tex]3x^2 + 15x + 18[/tex] which is the same as when you expand (3x + 6)(x + 3).

Answer:

(3x+9)(x+2)

step:3xx+3x2+9x+9x2

3x2+6x+9x+9x2

answer is 3x2+15x+18

step:3xx+3x3+6x+6x3

3x2+6x+6x3

answer is 3x2+15x+18

Step-by-step explanation:

#Carry on learning#

Hey there!

4 quarters = 1 dollar

To find how many quarters are in 20 dollars, we multiply 4 by 20

⇒ 4 × 20

⇒ 80

Therefore, 80 quarters are in 20 dollars

Answer:

See below for answers and explanations

Step-by-step explanation:

Problem 1

[tex]x=-3+2cos\theta,\:y=5+2sin\theta\\\\x+3=2cos\theta,\: y-5=2sin\theta\\\\(x+3)^2=4cos^2\theta,\: (y-5)^2=4sin^2\theta\\\\(x+3)^2+(y-5)^2=4cos^2\theta+4sin^2\theta\\\\(x+3)^2+(y-5)^2=4(cos^2\theta+sin^2\theta)\\\\(x+3)^2+(y-5)^2=4(1)\\\\(x+3)^2+(y-5)^2=4[/tex]

Thus, the first option is correct. Trying all the other options will not get you the desired rectangular equation.

Problem 2

[tex]x=3-6cos\theta,\: y=-2+3sin\theta\\\\x-3=-6cos\theta,\: y+2=3sin\theta\\\\\frac{x-3}{-6}=cos\theta,\: \frac{y+2}{3}=sin\theta\\ \\ \frac{(x-3)^2}{36}=cos^2\theta,\: \frac{(y+2)^2}{9}=sin^2\theta\\ \\ \frac{(x-3)^2}{36}+\frac{(y+2)^2}{9}=cos^2\theta+sin^2\theta\\\\ \frac{(x-3)^2}{36}+\frac{(y+2)^2}{9}=1[/tex]

Therefore, the first option is correct. This equation is in the form of an ellipse with a horizontal major axis length of 12 (half is 6) and a vertical minor axis length of 6 (half is 3), with its center at (3,-2).

Problem 3

Not sure which equation needs to be used for this problem

Problem 4

[tex]x=-7cos\theta ,\:y=5sin\theta\\\\-\frac{x}{7}=cos\theta,\: \frac{y}{5}=sin\theta\\ \\ \frac{x^2}{49}=cos^2\theta,\: \frac{y^2}{25}=sin^2\theta\\ \\\frac{x^2}{49}+\frac{y^2}{25}=cos^2\theta+sin^2\theta\\ \\ \frac{x^2}{49}+\frac{y^2}{25}=1[/tex]

This equation is in the form of an ellipse with a horizontal major axis length of 14 (half is 7) and a vertical minor axis length of 10 (half is 5). See attached graph.

Problem 5

Eliminate the parameter:

[tex]x=-t^2-2,\:y=-t^3+4t\\\\x+2=-t^2\\\\-x-2=t^2\\\\\pm\sqrt{-x-2}=t\\\\y=-t^3+4t\\\\y=-(\pm\sqrt{-x-2})^3+4(\pm\sqrt{-x-2})[/tex]

Attached below is the graph of the curve, which corresponds with the first option.