please solve WILL MARK BRAINLIES
Solve the question 2 it is necessary but 1 upto you​

Answer:

Step-by-step explanation:

irrational number

Answer:

Irrational number

Step-by-step explanation:

It is a never-ending and non-repeating number.

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:

y + 1 = -1/4 (x - 1) Answer choice C is correct.

Step-by-step explanation:

Point Slope Formula: y - y1 = m (x-x1)

Your points:

(1,-1) and (5,-2)

You need to find the slope first:

Use the formula: y2 - y1 / x2 - x1

Your y2 is -2

y1 is -1

x2 is 5

x1 is 1

-2 - (-1)/5-1

-2 + 1 /4

-1/4 is your slope and the "m" in the formula.

Now we know our y1 is -1 and x1 is 1 you just need to plug them in

y + 1 = -1/4 (x - 1)

Notice that I didn't write y - (-1) this is because the negatives cancel into positives.

Answer choice C is correct.

Answer:

c. [tex]y+1=-\frac{1}{4} (x-1)[/tex]

Step-by-step explanation:

Hi there!

We are given the points (1, -1) and (5, -2)

We want to find the equation of that line using those points, in point-slope form

Point-slope form is written as [tex]y-y_1=m(x-x_1)[/tex], where m is the slope and [tex](x_1, y_1)[/tex] is a point

First, let's find the slope of the line

The formula for the slope (m) calculated from 2 points is [tex]\frac{y_2-y_1}{x_2-x_1}[/tex], where [tex](x_1, y_1)[/tex] and [tex](x_2, y_2)[/tex] are points

We already have everything we need to find the slope, but let's label the values of the points to avoid any confusion when calculating.

[tex]x_1=1\\y_1=-1\\x_2=5\\y_2=-2[/tex]

Now substitute:

m=[tex]\frac{y_2-y_1}{x_2-x_1}[/tex]

m=[tex]\frac{-2--1}{5-1}[/tex]

Subtract

m=[tex]\frac{-2+1}{5-1}[/tex]

m=[tex]\frac{-1}{4}[/tex]

The slope of the line is -1/4

Now substitute this into the formula to find point-slope form (remember that this is [tex]y-y_1=m(x-x_1)[/tex], and that m is the slope value)

Therefore:

[tex]y-y_1=-\frac{1}{4} (x-x_1)[/tex]

Now, let's substitute the values of [tex]x_1[/tex] and [tex]y_1[/tex], which we found earlier (which are 1 and -1 respectively) into the equation

[tex]y--1=-\frac{1}{4} (x-1)[/tex]

Simplify

[tex]y+1=-\frac{1}{4} (x-1)[/tex]

This equation matches option c, which is the answer.
Hope this helps!

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.