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Vinvika [58]
3 years ago
11

How can you use the quadratic formula to complete a square?

Mathematics
2 answers:
Marina86 [1]3 years ago
6 0
It has become somewhat fashionable to have students derive the Quadratic Formula themselves; this is done by completing the square for the generic quadratic equation ax2 + bx + c = 0. While I can understand the impulse (showing students how the Formula was invented, and thereby providing a concrete example of the usefulness of abstract symbolic manipulation), the computations involved are often a bit beyond the average student at this point.
notka56 [123]3 years ago
5 0
<span><span>Step 1 Divide all terms by a (the coefficient of x2).</span>Step 2 Move the number term (c/a) to the right side of the equation.<span>Step 3 Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation.</span></span>
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A sequence of numbers is formed by the iterative process Xn+1 = (Xn )2 – Xn
xeze [42]

Step-by-step explanation:

X1 = 1

X2 = (X1) 2 - X1

X2 = 2 - 1

X2 = 1. implies that the second term of the sequence is 1.

3 0
2 years ago
Triangles ABC and DEF are similar. what is the length of side EF
Leto [7]
The anwser is either 2.88 or 3, But others are saying it is 3.
8 0
3 years ago
Read 2 more answers
Say that you own property with a market price of $74,000. The state tax assessors have given it an assessed value of 48% of that
BigorU [14]

Answer:

A

Step-by-step explanation:

74,000 x .48 = 35520. 35520 is the amount of property you have to pay for.

35520 x 0.031 = 1,101.12.  1.101.12 is the amount you owe in property tax.

5 0
3 years ago
Solve the following absolute value equations. Show the solution set and check your answers. |0.3-3/5k|-0.4=1.2
9966 [12]

Answer:

**The equation is not clear, so I have provided both options**

<h3><u>Option 1</u></h3>

\left|0.3-\dfrac{3}{5}k\right|-0.4=1.2

\implies \left|0.3-\dfrac{3}{5}k\right|=1.6

<u>Solution 1</u>

\implies 0.3-\dfrac{3}{5}k=1.6

\implies -\dfrac{3}{5}k=1.3

\implies k=-\dfrac{13}{6}

<u>Solution 2</u>

\implies -(0.3-\dfrac{3}{5}k)=1.6

\implies -0.3+\dfrac{3}{5}k=1.6

\implies \dfrac{3}{5}k=1.9

\implies k=\dfrac{19}{6}

<h3><u>Option 2</u></h3>

\left|0.3-\dfrac{3}{5k}\right|-0.4=1.2

\implies \left|0.3-\dfrac{3}{5k}\right|=1.6

<u>Solution 1</u>

\implies 0.3-\dfrac{3}{5k}=1.6

\implies -\dfrac{3}{5k}=1.3

\implies -3=6.5k

\implies k=-\dfrac{6}{13}

<u>Solution 2</u>

\implies -(0.3-\dfrac{3}{5k})=1.6

\implies -0.3+\dfrac{3}{5k}=1.6

\implies \dfrac{3}{5k}=1.9

\implies 3=9.5k

\implies k=\dfrac{6}{19}

6 0
2 years ago
Only need help with #'s 10, 12, 14, 16
gregori [183]
Im just about to work one out for you and you just do the same on the rest






7 0
3 years ago
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