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ladessa [460]
3 years ago
11

Which graph decreases, crosses the y-axis at (0,-7), and then remains constant?

Mathematics
1 answer:
sergey [27]3 years ago
7 0

Answer:

A.Graph B

Step-by-step explanation:

it decreases to a y of (0,-7) and then stays constant at -7

Hope this was helpful

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Volume of a cylinder of radius r and heigth h is
\pi {r}^{2} h
Here h is 15 cm and r is 6 cm.
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3 years ago
Write an equation of a line that is perpendicular to the line y=2/3x and passes through origin
sesenic [268]

keeping in mind that perpendicular lines have negative reciprocal slopes, hmmm what's the slope of the equation above anyway?

\bf y = \cfrac{2}{3}x\implies y = \stackrel{\stackrel{m}{\downarrow }}{\cfrac{2}{3}}x+0\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill

\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{2}{3}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{3}{2}}\qquad \stackrel{negative~reciprocal}{-\cfrac{3}{2}}}

so we're really looking for the equation of a line whose slope is -3/2 and runs through (0,0).

\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{0})~\hspace{10em} \stackrel{slope}{m}\implies -\cfrac{3}{2} \\\\\\ \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-\stackrel{y_1}{0}=\stackrel{m}{-\cfrac{3}{2}}(x-\stackrel{x_1}{0})\implies y=-\cfrac{3}{2}x

7 0
3 years ago
A community swimming pool is a rectangular prism that is 30 feet long, 12 feet wide, and 5 feet deep. The wading pool is half as
topjm [15]

Answer:

The volume of the community swimming pool is 4 times greaters than the volume of the wading pool.

Step-by-step explanation:

By definition of rectangular prism, we get the respective formulas for the volumes of the community swimming pool and the wadling pool, respectively:

Community swimming pool

V_{c} = l\cdot w\cdot h (1)

Wading pool

V_{w} = \left(\frac{1}{2}\cdot l \right)\cdot \left(\frac{1}{2}\cdot h\right)\cdot w (2)

Where:

l – Length of the swimming pool, measured in feet.

h – Depth of the swimming pool, measured in feet.

w – Width of the swimming pool, measured in feet.

V_{c} – Volume of the community swimming pool, measured in cubic feet.

V_{w} – Volume of the wading swimming pool, measured in cubic feet.

The ratio of the volume of the community swimming pool to the volume of the wadling pool is:

\frac{V_{c}}{V_{w}} = \frac{l\cdot w \cdot h}{\left(\frac{1}{2}\cdot l \right)\cdot \left(\frac{1}{2}\cdot h\right)\cdot w} (3)

\frac{V_{c}}{V_{w}} = \frac{1}{\frac{1}{4} }

\frac{V_{c}}{V_{w}} = 4

The volume of the community swimming pool is 4 times greaters than the volume of the wading pool.

8 0
2 years ago
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Math help
inn [45]
You should get photo math you can take a picture and it will give you the answer and show you how to solve it
8 0
3 years ago
What is the value of fraction 1 over 2x3 + 5.3y, when x = 2 and y = 3?
NikAS [45]

Hello, let's figure this one out together.

Me, myself had to write this down in the notes form to understand what I was doing.

In the fraction, multiply the 2's by 3.

1 over 2(2^3)+5.3(3)

1 over 2(2^3)+(5.3)(3)

Simplify each by step.

You end of up with a decimal.

You have a total of 0.031348.

4 0
3 years ago
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