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DanielleElmas [232]
3 years ago
10

In the polygon pictured, which angles are supplementary? [[ ANSWER ASAP ]]

Mathematics
1 answer:
denis-greek [22]3 years ago
5 0

Answer:

G and f

Step-by-step explanation:

The two angles next to each other make up 180 degrees.

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2nd one is the answer to it.
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3 years ago
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A model rocket is launched with an initial upward velocity of 60 m/s. The rocket's height h (in meters) after t seconds is given
Vedmedyk [2.9K]

Answer:

0.47 and 11.53

Step-by-step explanation:

h = 60t − 5t²

27 = 60t − 5t²

5t² − 60t + 27 = 0

Quadratic formula:

x = [ -b ± √(b² − 4ac) ] / 2a

t = [ -(-60) ± √((-60)² − 4(5)(27)) ] / 2(5)

t = (60 ± √3060) / 10

t = 0.47 or 11.53

3 0
3 years ago
Find the length and width of a rectangle that has the given area and a minimum perimeter. Area: 162 square feet
Gnom [1K]

Answer:

The width and length of rectangle is 12.728 m

Step-by-step explanation:

Let the length of the rectangle = L

let the width of the rectangle = W

The subjective function is given by;

F(p) = 2(L + W)

F = 2L + 2W

Area of the rectangle is given by;

A = LW

LW = 162 ft²

L = 162 / W

Substitute in the value of L into subjective function;

f = 2l + 2w\\\\f = 2(\frac{162}{w} )+2w\\\\f = \frac{324}{w} + 2w\\\\\frac{df}{dw} = \frac{-324}{w^2} +2\\\\

Take the second derivative of the function, to check if it will given a minimum perimeter

\frac{d^2f}{dw^2}= \frac{648}{w^3} \\\\Thus, \frac{d^2f}{dw^2}>0, \ since,\frac{648}{w^3} >0 \ (minimum \ function \ verified)

Determine the critical points of the first derivative;

df/dw = 0

\frac{-324}{w^2} +2 = 0\\\\-324 + 2w^2=0\\\\2w^2 = 324\\\\w^2 = \frac{324}{2} \\\\w^2 = 162\\\\w= \sqrt{162}\\\\w = 12.728 \ m

L = 162 / 12.728

L = 12.728 m

Therefore, the width and length of rectangle is 12.728 m

3 0
3 years ago
in a study by giacobbi et al. (2014), it was determined that about 60% of the participants found imagery very helpful in enhanci
Vanyuwa [196]

Before offering imaging programs, practitioners need to be aware of the reasons why exercise participants are doing it.

<h3>Define imagery in sports.</h3>

When we use imagery, we simulate an actual situation in our minds rather than actually going through it. It differs significantly from daydreaming or simply thinking about anything because it is a cognitive activity that is consciously used by an athlete or exerciser to accomplish a certain task.

In this study, an analysis of secondary data from a recently published randomized controlled trial. In a community-based, group-mediated physical activity intervention for sedentary people 50 and older, the Active Adult Mentoring Program (AAMP) tested the effectiveness of peer volunteers as delivery agents. The AAMP was built on the social-cognitive and self-determination theories, and mentors were trained to lead discussions in groups that would help reinforce key ideas from both theories.

The adaptability of images makes it useful at different times and in varied settings. Athletes employ imagery most frequently right before a competition or during practice, but they do so during the entire season, including the off-season. Similar to how it's reported by athletes, visualization is frequently used before an activity session. For example, it would be more effective for a swimmer to mentally practice her race start by adopting the proper position on the starting block at the swimming pool, as opposed to sitting on a chair at home. Both types of people will typically imagine within the sport and exercise environment where the benefits of this technique are maximized.

To know more about imagery in sports visit:

brainly.com/question/14319340

#SPJ4

3 0
1 year ago
What is the radius of a sphere with a volume of 1/6 pie
Yakvenalex [24]

Answer:

\large\boxed{the\ radius\ R=\dfrac{1}{2}}

Step-by-step explanation:

The formula of a volume of a sphere:

V=\dfrac{4}{3}\pi R^3

We have

V=\dfrac{1}{6}\pi

Substitute:

\dfrac{4}{3}\pi R^3=\dfrac{1}{6}\pi        <em>divide both sides by π</em>

\dfrac{4}{3}R^3=\dfrac{1}{6}          <em>multiply both sides by 3</em>

3\!\!\!\!\diagup^1\cdot\dfrac{4}{3\!\!\!\!\diagup_1}R^3=3\!\!\!\!\diagup^1\cdot\dfrac{1}{6\!\!\!\!\diagup_2}

4R^3=\dfrac{1}{2}          <em>divide both sides by 4</em>

R^3=\dfrac{1}{2}:4\\\\R^3=\dfrac{1}{2}\cdot\dfrac{1}{4}\\\\R^3=\dfrac{1}{8}\to R=\sqrt[3]{\dfrac{1}{8}}\\\\R=\dfrac{\sqrt1}{\sqrt8}\\\\R=\dfrac{1}{2}

3 0
3 years ago
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