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JulsSmile [24]
3 years ago
5

a soccer team practiced 3 times a week. each practice lasts for 110 minutes. the team practices for 14 weeks during the season.

how many minutes did he team practice during the season? how many hours did the team practice during the season? ( Hint: remember that there are 60 minutes in one hour)
Mathematics
1 answer:
Fittoniya [83]3 years ago
5 0
3 weeks x 110 minutes = 330 minutes
330 minutes a week x 14 weeks = 4620 minutes
4620 divided by 60 minutes = 77 hours
77 hours is your answer
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Step-by-step explanation:

OK, let's assume it this way:

<em>Sn=1.1!+2.2!+3.3!+...+n.n!</em><em>=</em><em>(</em><em>2</em><em>‐</em><em>1</em><em>)</em><em>.</em><em>1</em><em>!</em><em>+</em><em>(</em><em>3</em><em>-</em><em>1</em><em>)</em><em>.</em><em>2</em><em>!</em><em>+</em><em>(</em><em>4</em><em>-</em><em>1</em><em>)</em><em>3</em><em>!</em><em>+</em><em>.</em><em>.</em><em>.</em><em>+</em><em>(</em><em>(</em><em>n</em><em>+</em><em>1</em><em>)</em><em>-</em><em>1</em><em>)</em><em>.</em><em>n</em><em>!</em>

Sn=1.1!+2.2!+3.3!+...+n.n!=(2‐1).1!+(3-1).2!+(4-1)3!+...+((n+1)-1).n!<em>=</em><em>(</em><em>2</em><em>.</em><em>1</em><em>!</em><em>-</em><em>1</em><em>!</em><em>)</em><em>+</em><em>(</em><em>3</em><em>.</em><em>2</em><em>!</em><em>-</em><em>2</em><em>!</em><em>)</em><em>+</em><em>(</em><em>4</em><em>.</em><em>3</em><em>!</em><em>-</em><em>3</em><em>!</em><em>)</em><em>+</em><em>.</em><em>.</em><em>.</em><em>+</em><em>(</em><em>(</em><em>n-1</em><em>)</em><em>n</em><em>!</em><em>-</em><em>n</em><em>!</em><em>)</em><em>=</em><em>(</em><em>2</em><em>!</em><em>-</em><em>1</em><em>!</em><em>)</em><em>+</em><em>(</em><em>3</em><em>!</em><em>-</em><em>2</em><em>!</em><em>)</em><em>+</em><em>(</em><em>4</em><em>!</em><em>-</em><em>3</em><em>!</em><em>)</em><em>+</em>

Sn=1.1!+2.2!+3.3!+...+n.n!=(2‐1).1!+(3-1).2!+(4-1)3!+...+((n+1)-1).n!=(2.1!-1!)+(3.2!-2!)+(4.3!-3!)+...+((n-1)n!-n!)=(2!-1!)+(3!-2!)+(4!-3!)+<em>.</em><em>.</em><em>.</em><em>+</em><em>(</em><em>n</em><em>+</em><em>1</em><em>)</em><em>!</em><em>-</em><em>n</em><em>!</em><em>=</em><em>(</em><em>n</em><em>+</em><em>1</em><em>)</em><em>!</em><em>-</em><em>1</em><em>!</em><em>=</em><em>(</em><em>n</em><em>+</em><em>1</em><em>)</em><em>!</em><em>-</em><em>1</em>

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Answer:

Therefore, the solutions of the quadratic equations are:

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The graph is also attached.

Step-by-step explanation:

The solution of the graph could be obtained by finding the x-intercept.

y=\left(x-\:4\right)^2-1

Finding the x-intercept by substituting the value y = 0

so

y=\left(x-\:4\right)^2-1

\:0\:=\:\left(x\:-\:4\right)^2\:-\:1        ∵ y = 0

\left(x-4\right)^2-1=0

\left(x-4\right)^2-1+1=0+1

\left(x-4\right)^2=1

\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

\mathrm{Solve\:}\:x-4=\sqrt{1}

\mathrm{Apply\:rule}\:\sqrt{1}=1

x-4=1

x=5

\mathrm{Solve\:}\:x-4=-\sqrt{1}

\mathrm{Apply\:rule}\:\sqrt{1}=1

x-4=-1

x=3

So, when y = 0, then x values are 3, and 5.

Therefore, the solutions of the quadratic equations are:

x=5,\:x=3

The graph is also attached. As the graph is a Parabola. It is visible from the graph that the values of y = 0 at x = 5 and x = 3. As the graph is a Parabola.

8 0
3 years ago
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