All categories

2 years ago

Rewrite the equation by completing the square. x^{2}-14x+33 = 0

Mathematics

2 answers:

makvit [3.9K]2 years ago

6 0

Answer:

Step-by-step explanation:

x^2 - 14x = -33

14/2= 7

(7)^2 = 49

x^2 - 14x + 49 = -33 + 49

x^2 -14x + 49 = 16

(x-7)^2 - 16

Alina [70]2 years ago

3 0

Answer:

x = 11

x =3

Step-by-step explanation:

I hope it helps

You might be interested in

PLEASE HELP ITS TIMED WILL GIVE MORE THEN 5 POINTS AND BRAINLIEST

Fofino [41]

Answer:

I believe it is c

Step-by-step explanation:

Correct me if I'm wrong but since the two lines look the same length I believe it is c

6 0

2 years ago

Find the equation of a line that is perpendicular to y = 3x – 5 and passes through the point (1, -3).

Svetradugi [14.3K]

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

$\begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}\qquad \qquad y = \stackrel{\stackrel{m}{\downarrow }}{3}x-5$

well then therefore

$\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{3\implies \cfrac{3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{3}}}$

so we're really looking for the equation of a line with slope of -1/3 and that passes through (1, -3 )

$(\stackrel{x_1}{1}~,~\stackrel{y_1}{-3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{-\cfrac{1}{3}}(x-\stackrel{x_1}{1})\implies y+3=-\cfrac{1}{3}x+\cfrac{1}{3} \\\\\\ y=-\cfrac{1}{3}x+\cfrac{1}{3}-3\implies y=-\cfrac{1}{3}x-\cfrac{8}{3}$

6 0

2 years ago

Please answer for me :

Arisa [49]

Answer:

Part A - $2\frac{1}{2} + b = 4\frac{2}{3}$

Part B - $2\frac{1}{6}$

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Express p in terms of q if

Irina-Kira [14]

$p=x^2+\dfrac{1}{x^2}\\\\p=\dfrac{x^4}{x^2}+\dfrac{1}{x^2}\\\\p=\dfrac{x^4+1}{x^2}\qquad(*)$

$q=x+\dfrac{1}{x}\\\\q=\dfrac{x^2}{x}+\dfrac{1}{x}\\\\q=\dfrac{x^2+1}{x}\qquad\text{square both sides}\\\\q^2=\left(\dfrac{x^2+1}{x}\right)^2\\\\q^2=\dfrac{(x^2+1)^2}{x^2}\qquad\text{use}\ \ (a+b)^2=a^2+2ab+b^2\\\\q^2=\dfrac{(x^2)^2+2(x^2)(1)+1^2}{x^2}\\\\q^2=\dfrac{x^4+2x^2+1}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+\dfrac{2x^2}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+2\qquad\text{subtract 2 from both sides}\\\\q^2-2=\dfrac{x^4+1}{x^2}\\\\\text{From (*) we have}\\\\\boxed{p=q^2-2}$

8 0

2 years ago

Solve the equation.<br><br> –9x + 1 = –x + 17

VikaD [51]

Answer:

-9x+1 = -x +17

-8x = 16

x = -2

Step-by-step explanation:

8 0

3 years ago