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Len
3 months ago
11

Two of the steps in the derivation of the quadratic formula are shown below. Step 6: StartFraction b squared minus 4 a c Over 4

a squared EndFraction = (x + StartFraction b Over 2 a EndFraction) squared Step 7: StartFraction plus or minus StartRoot b squared minus 4 a c EndRoot Over 1 a EndFraction = x + StartFraction b Over 2 a EndFraction Which operation is performed in the derivation of the quadratic formula moving from Step 6 to Step 7? subtracting StartFraction b Over 2 a EndFraction from both sides of the equation squaring both sides of the equation taking the square root of both sides of the equation taking the square root of the discriminant

Mathematics
2 answers:
babunello [11.8K]3 months ago
6 0

Explanation:

Step-by-step clarification:

Referring to step 6

(b² — 4ac) / 4a² = (x + b/2a)²

The mistake in the question is that it should be (x + b/2a)²

According to step 7

±√(b² —4ac) /2a = x + b/2a

The error in the question is that it should be divided by 2a, not 1a.

1. The transition from step 6 to step 7 involves taking the square roots of both sides

(b² — 4ac) / 4a² = (x + b/2a)²

Taking the square of both sides

√(b²—4ac) / √4a² = √(x + b/2a)²

√(b²—4ac) / 2a = x + b/2a

This forms step 7 correctly.

Next, subtracting b/2a from both sides

√(b²—4ac) / 2a - b/2a= x + b/2a -b/2a

√(b²—4ac) / 2a — b/2a = x

(√(b²—4ac)  — b)/2a = x

x = [—b ± √(b²—4ac)] / 2a

This gives the desired formula.

The discriminant is D = b²—4ac.

Svet_ta [12.7K]3 months ago
5 0

Explanation:

c

Detailed reasoning:

Square roots of both sides are extracted from the equation

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When the sun is at a certain angle in the sky, a 50-foot building will cast a -foot shadow. What is the length of the shadow in
Svet_ta [12734]

Answer:

0.40 feet

Step-by-step explanation:

For the first scenario, a 50-foot building casts a shadow of 1 foot. Let the angle of elevation of the sun from the shadow be denoted as θ.

Then:

Tan θ = \frac{opposite}{adjacent}

Tan θ = \frac{50}{1}

Tan θ = 50

⇒ θ = Tan^{-1} 50

      = 88.8542

      = 88.85^{o}

The elevation angle is roughly 88.85^{o}.

For a 20-foot pole,

Tan θ = \frac{opposite}{adjacent}

Tan 88.85^{o} = \frac{20}{x}

x = \frac{20}{Tan 88.85^{o} }

 = 0.4015

 = 0.40 feet

The length of the pole's shadow is 0.40 feet.

4 0
2 months ago
What is a3 in an arithmetic sequence in which a10=41 and a15=61
zzz [12365]
\bf \begin{array}{llll}
term&value\\
-----&-----\\
a_{10}&41\\
a_{11}&41+d\\
a_{12}&(41+d)+d\\
&41+2d\\
a_{13}&(41+2d)+d\\
&41+3d\\
a_{14}&(41+3d)+d\\
&41+4d\\
a_{15}&(41+4d)+d\\
&41+5d=61
\end{array}
\\\\\\
41+5d=61\implies 5d=20\implies d=\cfrac{20}{5}\implies \boxed{d=4}\\\\
-------------------------------\\\\

\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
d=4\\
n=10\\
a_{10}=41
\end{cases}
\\\\\\
41=a_1+(10-1)4\implies 41=a_1+36\implies \boxed{5=a_1}

therefore

\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
d=4\\
n=3\\
a_{1}=5
\end{cases}
\\\\\\
a_3=a_1+(3-1)4\implies a_3=5+(3-1)4

and you are probably aware of the amount.
8 0
2 months ago
Read 2 more answers
John runs 500 feet in 1 minute. Identify the correct conversion factor setup required to compute John's speed in inches per seco
PIT_PIT [12445]

Answer:

refer to the method

Step-by-step explanation:

Keep in mind that

1\ ft=12\ in

To change feet to inches, multiply by 12

1\ min=60\ sec

To convert minutes into seconds, multiply by 60

we have

500\frac{ft}{min}=500(\frac{12}{60})=500(\frac{1}{5})=100\frac{in}{sec}

7 0
3 months ago
Read 2 more answers
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