A triangular window is above the door to a café. The length of the base of the window is 12 feet, and the height is 10 feet.



What is the area of the window?

Enter your answer in the box.

ft2

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:

Step-by-step explanation:

irrational number

Answer:

Irrational number

Step-by-step explanation:

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

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.

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#

Answer:

1.4.5m/s^2

2.6.25m/s^2

3.15m/s^2

4.0.05m/s^2

Step-by-step explanation:

a=v-u

----

t

27-0

------ = 27/6

6

=4.5m/s^2

u=4.5m/s

v=24.5m/s

t=3.2s

(24.5-4.5)÷3.2

=20/3.2

=6.25m/s^2

v=80m/s

u=50m/s

t=2s

(80-50)÷2

15m/s^2

v=0.80m/s

u=0.50m/s

t=6s

(0.80-0.50)÷6

=0.05m/s^2

Answer:

  B.) y=csc x

Step-by-step explanation:

The cosecant function has a minimum magnitude of 1, so its range excludes any values in the range -1 < y < 1.

  y = csc(x) . . . has a range that does not include -0.8