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TEA [102]
3 years ago
12

An investor invests $500 in a certain stock. After the first six months, the value if the stock has increased at a rate of $20 p

er month. Sketch a graph that represents the situation.

Mathematics
2 answers:
seraphim [82]3 years ago
6 0

Answer:

We know that:

  • Total investment is $500.
  • The stock give $20 per month.

Based on this, we can make a table and then graph. <em>x </em>variable is gonna be months, and <em>y </em>variable is gonna be the money. We just need to write six months, and each month increase $20. The initial condition is <em>x=0</em>, in this moment, the money is $500.

x      y

0     500

1      520

2     540

3     560

4     580

5     600

6     620

Remember that each (x;y) pair represents a point to graph the function.

stiv31 [10]3 years ago
3 0
I cannot draw a graph so im going to explain 
on the bottom u write per month and on the side and then increase the amount of money by 20$.
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Step-by-step explanation:

The student made a mistake by identifying the maximum point by the x coordinate value of the vertex. Minimum or maximum points are determine using the y coordinate value.The student can determine the difference between a maximum and minimum by identifying the y coordinate of the vertex.

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If there is one-third of a pizza left, and Becky only wants to eat one-half of it, how much of the pizza does she want?
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She wants to eat 1/6 of the pizza.
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2 years ago
Read 2 more answers
A teacher places n seats to form the back row of a classroom layout. Each successive row contains two fewer seats than the prece
Alex_Xolod [135]

Answer:

The number of seat when n is odd S_n=\frac{n^2+2n+1}{4}

The number of seat when n is even S_n=\frac{n^2+2n}{4}

Step-by-step explanation:

Given that, each successive row contains two fewer seats than the preceding row.

Formula:

The sum n terms of an A.P series is

S_n=\frac{n}{2}[2a+(n-1)d]

    =\frac{n}{2}[a+l]

a = first term of the series.

d= common difference.

n= number of term

l= last term

n^{th} term of a A.P series is

T_n=a+(n-1)d

n is odd:

n,n-2,n-4,........,5,3,1

Or we can write 1,3,5,.....,n-4,n-2,n

Here a= 1 and d = second term- first term = 3-1=2

Let t^{th} of the series is n.

T_n=a+(n-1)d

Here T_n=n, n=t, a=1 and d=2

n=1+(t-1)2

⇒(t-1)2=n-1

\Rightarrow t-1=\frac{n-1}{2}

\Rightarrow t = \frac{n-1}{2}+1

\Rightarrow t = \frac{n-1+2}{2}

\Rightarrow t = \frac{n+1}{2}

Last term l= n,, the number of term =\frac{ n+1}2, First term = 1

Total number of seat

S_n=\frac{\frac{n+1}{2}}{2}[1+n}]

    =\frac{{n+1}}{4}[1+n}]

     =\frac{(1+n)^2}{4}

    =\frac{n^2+2n+1}{4}

n is even:

n,n-2,n-4,.......,4,2

Or we can write

2,4,.......,n-4,n-2,n

Here a= 2 and d = second term- first term = 4-2=2

Let t^{th} of the series is n.

T_n=a+(n-1)d

Here T_n=n, n=t, a=2 and d=2

n=2+(t-1)2

⇒(t-1)2=n-2

\Rightarrow t-1=\frac{n-2}{2}

\Rightarrow t = \frac{n-2}{2}+1

\Rightarrow t = \frac{n-2+2}{2}

\Rightarrow t = \frac{n}{2}

Last term l= n, the number of term =\frac n2, First term = 2

Total number of seat

S_n=\frac{\frac{n}{2}}{2}[2+n}]

    =\frac{{n}}{4}[2+n}]

     =\frac{n(2+n)}{4}

    =\frac{n^2+2n}{4}  

4 0
3 years ago
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Tanya [424]

second side = x

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x = 12 m

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(3x - 6) + 8 = 3x + 2 = 3(12) + 2 = 38 m

Answer: 12m, 30m, 38m

8 0
3 years ago
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