A quadratic equation is shown:x2 - 14x + 41 = 0
Which of the following is the first correct step to write the above equation in the form (x - p)2 = 9, where p and q are integers?
x2 - 14x + 41 + 8 = 0 + 8
x2 - 14x + 41 + 9 = 0 + 9
x2 - 14x + 41 - 8 = 0 - 8
x2 – 14x + 41 - 9 = 0 - 9

Answers

Answer 1
Answer:

Answer:

The correct answer is the first option

Step-by-step explanation:

Quadratic Equation

The standard form of a quadratic equation is

ax^2+bx+c=0

Sometimes we need to change the expression of the same equation to the form

(x-p)^2=q

To accomplish that change, we usually modify the left-hand expression to make it look like the square of a binomial.

The given quadratic equation is

x^2-14x+41=0

Recall the square of a binomial is

(x-p)^2=x^2-2px+p^2

The first term is already present. The second term gives us the value of p:

-2px=-14x

Solving

p=7

Now we need to produce the third term p^2=49. We only have 41, thus we need to add 8 to both sides of the equation:

x^2-14x+41+8=0+8

The correct answer is the first option


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Which lst of ordered pairs represents solutions to x+y= 2? (4.6), (0, 2), (4, 2) (4,6), (0, 2). (4, 2) (4,-6), (0, 2), (4, 2) ​

Answers

answer: x+y=2

0+2=²

( 0 , 2 )

Answer:

(0,2)

Step-by-step explanation:

0 + 2 = 2

Where x = 0, y = 2

quiz 4-1 : classifying and solving for sides/angles in triangles gina wilson (all things algebra) URGENT

Answers

Answer:

14.

x = 11  m<B = 108   m<C= 36   m<D = 36

15.

x = 4     WX = 25     XY = 17     WY = 25

Step-by-step explanation:

Solving (14):

BC = BD

m<B = 13x - 35

m<C = 5x - 19

m<D = 2x + 14

First, we solve for x.

Since sides BC = BD, then

m<C = m<D

This is so because the theorem of ASA (Angle-Side-Angle)

So, we have:

5x - 19 = 2x + 14

Collect Like Terms

5x - 2x = 19 + 14

3x = 33

Divide both sides by 3

x = 11

Solving for m<B, m<C and m<D.

We simply substitute 11 for x in their respective expressions.

m<B = 13x - 35

m<B = 13*11 - 35

m<B = 143 - 35

m<B = 108

m<C = 5x - 19

m<C= 5*11 - 19

m<C= 55 - 19

m<C= 36

m<D = 2x + 14

m<D = 2*11 + 14

m<D = 22 + 14

m<D = 36

Solving (15):

<X = <Y

WX = 9x - 11

XY = 4x + 1

WY = 7x - 3

First, we solve for x.

Since <X = <Y, then

WX = WY

This is so because the theorem of SAS (Side-Angle-Side)

So, we have:

9x - 11 = 7x - 3

Collect Like Terms

9x - 7x = 11-3

2x = 8

Divide both sides by 2

x = 4

Solving for WX, WY and XY

We simply substitute 4 for x in their respective expressions.

WX = 9x - 11

WX = 9 * 4 - 11

WX = 36 - 11

WX = 25

XY = 4x + 1

XY = 4 * 4 + 1

XY = 16 + 1

XY = 17

WY = 7x - 3

WY = 7 * 4 - 3

WY = 28 - 3

WY = 25

Help please whats the answer

Answers

Answer:

12 students

Explanation:

40% earned a B; 20% earned an A, so (40%+20% =) 60% got a "B or better."

60% of 20 students is ...

... 60/100 × 20 students = 1200/100 students = 12 students

Omg help me giving brainlyest!! plus 30 points

Answers

Answer:

5

Step-by-step explanation:

Answer:

Step-by-step explanation:

You have to use PEMDAS

first p, parenthesis

12-2=10

then e exponenets 3^3=9

and finally d diviison /2

so 9(10) : 2

=90/2

=45

Evaluate the given integral by changing to polar coordinates. sin(x2 y2) dA R , where R is the region in the first quadrant between the circles with center the origin and radii 2 and 4

Answers

Answer:

I = 1.47001

Step-by-step explanation:

we have the function

f(x,y)=sin(x^2y^2)\n

In polar coordinates we have

x=rcos\theta\ny=rsin\theta

and dA is given by

dA=rdrd\theta

Hence, the integral that we have to solve is

I=\int \limt_2^4 \int \limit_0^(\pi /2)sin(r^4cos^2\theta sin^2\theta)rdrd\theta

This integral can be solved in a convenient program of your choice (it is very difficult to solve in an analytical way, I use Wolfram Alpha on line)

I = 1.47001

Hope this helps!!!

Evaluate 3.2 + 6 to the second power minus 2 * 7.2 show your work will mark brainiest

Answers

3.2 + 6^2 - 2 * 7.2

3.2 + 36 - 2 * 7.2

3.2 + 36 - (-14.4)

3.2 + 36 + 14.4

24.8

24.8 is your answer.