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2 years ago

Pleaseeeee help (15 points)

Mathematics

1 answer:

Elina [12.6K]2 years ago

5 0

Answer:

Step-by-step explanation:

what are the vocabulary words that u are supposed to use?

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HELP HELP HELP HELP HELP HELP plsssssss¡!

mylen [45]

Answer:

$201.60

Step-by-step explanation:

12% of 180 = 21.6

Cost of health club = 180

180 + 21.6 = $201.60

7 0

2 years ago

Find an equation of the parabola that has the indicated vertex and whose graph passes through the given point

vichka [17]

Equation of a parabola with vertex at (2, -1) is
y = a(x - 2)^2 - 1

Using the given point: -3 = a(4 - 2)^2 - 1
-2 = a(2)^2
4a = -2
a = -1/2

Therefore, required equation is
y = -1/2(x - 2)^2 - 1
y = -1/2(x^2 - 4x + 4) - 1
y = -1/2x^2 + 2x - 2 - 1
y = -1/2x^2 + 2x - 3

3 0

2 years ago

Which of the equations below has a slope of 1/4

kenny6666 [7]

Answer:

uh you didnt put the picture please put the picture i really want to help you:)

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Find the diameter(s) of circle O.

schepotkina [342]

Answer:

The diameter of the circle O is <em>12 units</em>

Step-by-step explanation:

There is no data in the image provided. To better help you, I'm assuming we have an arbitrary value of

$BC=6\sqrt{2}$

and the triangle B0C is right

<u>Relations in the Circle</u>

The diameter (D) is twice the radius (r) and the radius is the distance measured from the center of the circle to any point of the circumference.

Since the triangle B0C has a right angle, BD is the hypotenuse and 0B=0C=r

Applying Pythagoras's theorem:

$BC^2=r^2+r^2=2r^2$

Thus we have

$2r^2=(6\sqrt{2})^2$

$2r^2=72$

$r^2=36$

$r=6$

The diameter of the circle O is D=2(6) = 12 units

5 0

3 years ago

A line passes through the points (2,4) and (5,6) . Select Yes or No to tell whether each equation describes this line. Equation

krek1111 [17]

$\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-4}{5-2}\implies \cfrac{2}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-4=\cfrac{2}{3}(x-2)$

3 0

2 years ago

Pleaseeeee help (15 points) ​

Pleaseeeee help (15 points)