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Amiraneli [1.4K]
3 years ago
7

If using the method of completing the square to solve the quadratic equation

Mathematics
2 answers:
lara [203]3 years ago
6 0

Answer:

x² + 7x + 10 = 0

Subtract 10 from both sides

x² + 7x = -10

Use half the x coefficent (7/2) as the complete the square term

(x + 7/2)² = -10 + (7/2)²

   note: the number added to "complete the square" is (7/2)² = 49/4

(x + 7/2)² = -10 + 49/4

(x + 7/2)² = 9/4

Take the square root of both sides

x + 7/2 = ±3/2

Subtract 7/2 from both sides

x = -7/2 ± 3/2

x = {-5, -2}

Katena32 [7]3 years ago
6 0
X2 cause it makes sense
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A positive number with an absolute value less than 5.
Goryan [66]

Answer:

3 I think.

Step-by-step explanation:

it's positive and it's also an absolute

4 0
3 years ago
How would you find the vertices of an image of a figure was rotated 270 degrees clockwise? 15 POINTS! DESPERATE
boyakko [2]

Rotation Coordinate Notation About the Origin:

90–degree counterclockwise: (x, y) → (-y, x)

180–degree counterclockwise: (x, y) → (-x, -y)

270–degree counterclockwise: (x, y) → (y, -x)

Use this to help you..

Example -

The vertices of a triangle are

( 1 , 3 ) ( 3 , -4 ) , ( - 6 , -8 )

270 degrees , clockwise , rotated from origin ( x , y ) - ( y , -x )

The first vertice = ( 1 , 3 )

after rotation - ( 3 , - 1 )

The second vertice = ( 3 , -4 )

After rotation - ( -4 , -3 )

The third vertice = ( -6 , - 8 )

After rotation - ( -8 , 6 )

Hope my answer helps!

5 0
4 years ago
Read 2 more answers
Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that
FromTheMoon [43]

Answer:

The Taylor series is \ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}.

The radius of convergence is R=3.

Step-by-step explanation:

<em>The Taylor expansion.</em>

Recall that as we want the Taylor series centered at a=3 its expression is given in powers of (x-3). With this in mind we need to do some transformations with the goal to obtain the asked Taylor series from the Taylor expansion of \ln(1+x).

Then,

\ln(x) = \ln(x-3+3) = \ln(3(\frac{x-3}{3} + 1 )) = \ln 3 + \ln(1 + \frac{x-3}{3}).

Now, in order to make a more compact notation write \frac{x-3}{3}=y. Thus, the above expression becomes

\ln(x) = \ln 3 + \ln(1+y).

Notice that, if x is very close from 3, then y is very close from 0. Then, we can use the Taylor expansion of the logarithm. Hence,  

\ln(x) = \ln 3 + \ln(1+y) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{y^n}{n}.

Now, substitute \frac{x-3}{3}=y in the previous equality. Thus,

\ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}.

<em>Radius of convergence.</em>

We find the radius of convergence with the Cauchy-Hadamard formula:

R^{-1} = \lim_{n\rightarrow\infty} \sqrt[n]{|a_n|},

Where a_n stands for the coefficients of the Taylor series and R for the radius of convergence.

In this case the coefficients of the Taylor series are

a_n = \frac{(-1)^{n+1}}{ n3^n}

and in consequence |a_n| = \frac{1}{3^nn}. Then,

\sqrt[n]{|a_n|} = \sqrt[n]{\frac{1}{3^nn}}

Applying the properties of roots

\sqrt[n]{|a_n|} = \frac{1}{3\sqrt[n]{n}}.

Hence,

R^{-1} = \lim_{n\rightarrow\infty} \frac{1}{3\sqrt[n]{n}} =\frac{1}{3}

Recall that

\lim_{n\rightarrow\infty} \sqrt[n]{n}=1.

So, as R^{-1}=\frac{1}{3} we get that R=3.

8 0
4 years ago
A circle with radius of 2cm sits inside a circle with radius of
Semmy [17]

Answer:

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Step-by-step explanation:

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area of circle with radius 2 cm = (pi)r^2 = 3.14159 * (2 cm)^2 = 12.57 cm^2

difference in areas = 37.70 cm^2

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The answer would be 0.4yd
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