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natka813 [3]
3 years ago
15

A man purchased 100 shares of a company’s stock at $3 per share. If each share rose $0.09 in value during the first month, decre

ased $0.02 during the second month, and gained $0.03 during the third month, what is the value, in dollars, of the man’s total investment at the end of the third month?
Mathematics
1 answer:
Neporo4naja [7]3 years ago
6 0
$310 dollars because after it res and decreased it gained $.10 and 100 x 3 is 300 so thats before you calculate how much it rose so .10 x 100 is 10 so add that to 300 and you get 310
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Find the first four terms of the sequence given by the following.<br> aₙ=43-3(n-1) n=1,2,3. . .
Serggg [28]

Given :-

  • The general term of a sequence is given by aₙ=43-3(n-1) .

To Find :-

  • The first four terms of the sequence.

Solution :-

The given expression is /

→ aₙ=43-3(n-1)

where n > 0

<u>Finding</u><u> the</u><u> </u><u>first </u><u>term </u><u>:</u>

Substituting n = 1 , we have ,

→ T1 = 43 - 3(1-1)

→ T1 = 43 - 3*0

→ T1 = 43 - 0 = 43

<u>Finding</u><u> the</u><u> </u><u>second</u><u> </u><u>term </u><u>:</u>

Substituting n = 2 , we have,

→ T2 = 43 -3(2-1)

→ T2 = 43 -3*1

→ T2 = 43 -3 = 40

<u>Finding</u><u> </u><u>the </u><u>third </u><u>term</u><u> </u><u>:</u>

Substituting n = 3 , we have,

→ T3 = 43 -3(3-1)

→ T3 = 43 -3*2

→ T3 = 43 -6 = 37

<u>Finding</u><u> the</u><u> </u><u>fourth</u><u> </u><u>term </u><u>:</u>

→ T4 = 43 -3(4-1)

→ T4 = 43 -3*3

→ T4 = 43-9 = 34

<u>Hence</u><u> the</u><u> </u><u>first</u><u> </u><u>four</u><u> terms</u><u> of</u><u> </u><u>the</u><u> </u><u>sequence</u><u> </u><u>are </u><u>4</u><u>3</u><u> </u><u>,</u><u> </u><u>4</u><u>0</u><u> </u><u>,</u><u> </u><u>37</u><u> </u><u>and </u><u>34</u><u> </u><u>.</u>

<em>I </em><em>hope</em><em> this</em><em> helps</em><em> </em><em>.</em><em> </em><em>Let </em><em>me</em><em> know</em><em> if</em><em> you</em><em> </em><em>need </em><em>further</em><em> </em><em>clarification</em><em> </em><em>.</em>

7 0
2 years ago
For what value of k are the roots of the quadratic equation kx (x-2)+6=0 equal?​
ser-zykov [4K]

Answer:

\boxed{\sf k=6}

Step-by-step explanation:

\sf kx (x-2)+6=0

Expand brackets.

\sf kx^2 -2kx+6=0

This is in quadratic form.

\sf ax^2 +bx+c=0

Since this is for equal roots:

\sf b^2 -4ac=0

\sf a=k\\b=-2k\\c=6

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\sf 4k(k-6)=0

\sf 4k=0\\k=0

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Plug k as 0 to check.

\sf \sf 0x^2 -2(0)x+6=0\\6=0

False.

So that means k must equal 6.

3 0
3 years ago
Number one solve sin l and tan n cos l and sin n
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Sin L= 3/5

Tan N= 3/5

Cos L=4/5

Sin N=4/5

5 0
3 years ago
8+0=18<br> Only answer<br><br> Plz someone very intelligence
Free_Kalibri [48]
I think
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just place 1 before 0
3 0
3 years ago
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Answer:

ok

Step-by-step explanation:

5'm

5 0
2 years ago
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