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vlabodo [156]
3 years ago
8

Use completing the square to solve for x in the equation (x-12)(x+4)= 9.

Mathematics
1 answer:
iragen [17]3 years ago
4 0

Answer:

x = 4 + sqrt(73) or x = 4 - sqrt(73)

Step-by-step explanation by completing the square:

Solve for x:

(x - 12) (x + 4) = 9

Expand out terms of the left hand side:

x^2 - 8 x - 48 = 9

Add 48 to both sides:

x^2 - 8 x = 57

Add 16 to both sides:

x^2 - 8 x + 16 = 73

Write the left hand side as a square:

(x - 4)^2 = 73

Take the square root of both sides:

x - 4 = sqrt(73) or x - 4 = -sqrt(73)

Add 4 to both sides:

x = 4 + sqrt(73) or x - 4 = -sqrt(73)

Add 4 to both sides:

Answer:  x = 4 + sqrt(73) or x = 4 - sqrt(73)

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$503.90

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What is the reason for number 4 and 5?
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3 years ago
1. In an auditorium, there are 21 seats in the first row and 26 seats in the second row. The number of seats in a row continues
kondor19780726 [428]
1. Let s_n be the number of seats in the n-th row. The number seats in the n-th row relative to the number of seats in the (n-1)-th row is given by the recursive rule

s_n=s_{n-1}+5


Since s_1=21, we have

s_2=s_1+5
s_3=s_2+5=s_1+2\cdot5
s_4=s_3+5=s_1+3\cdot5
\cdots
s_n=s_{n-1}+5=\cdots=s_1+(n-1)\cdot5

So the explicit rule for the sequence s_n is

s_n=21+5(n-1)\implies s_n=5n+16

In the 15th row, the number of seats is


s_{15}=5(15)+16=91

2. Let p_n be the amount of profit in the n-th year. If the profits increase by 6% each year, we would have

p_2=p_1+0.06p_1=1.06p_1
p_3=1.06p_2=1.06^2p_1
p_4=1.06p_3=1.06^3p_1
\cdots
p_n=1.06p_{n-1}=\cdots=1.06^{n-1}p_1

with p_1=40,000.

The second part of the question is somewhat vague - are we supposed to find the profits in the 20th year alone? the total profits in the first 20 years? I'll assume the first case, in which we would have a profit of


p_{20}=1.06^{19}\cdot40,000\approx121,024

3. Now let p_n denote the number of pushups done in the n-th week. Since 3\cdot4=12, 12\cdot4=48, and 48\cdot4=192, it looks like we can expect the number of pushups to quadruple per week. So,

p_n=4p_{n-1}

starting with p_1=3.

We can apply the same reason as in (2) to find the explicit rule for the sequence, which you'd find to be

p_n=4^{n-1}p_1\implies p_n=4^{n-1}\cdot3
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I agree, the only 2 that look the most similar is B. and D.
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