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vodka [1.7K]
3 years ago
14

What number must you add to complete the square x^2+xA. 1/2B. 2C. 1/4D. 1

Mathematics
1 answer:
Irina-Kira [14]3 years ago
8 0

for ax^2+bx+c

when a=1, then the value of c that will make the quadratic a perfect square (allows you to complete the square) is (\frac{b}{2})^2


given x^2+x

b=1

so c=(\frac{1}{2})^2=\frac{1}{4}

answer is C

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This one might not look familiar yet, but it's very similar to the distance problems you've already seen.
Debora [2.8K]

Answer:

176.

Step-by-step explanation:

5 0
3 years ago
Write an equation of the hyperbola given that the center is at (2, -3), the vertices are at (2, 3) and (2, - 9), and the foci ar
zavuch27 [327]
Check the picture below.

so, the hyperbola looks like so, clearly a = 6 from the traverse axis, and the "c" distance from the center to a focus has to be from -3±c, as aforementioned above, the tell-tale is that part, therefore, we can see that c = 2√(10).

because the hyperbola opens vertically, the fraction with the positive sign will be the one with the "y" in it, like you see it in the picture, so without further adieu,

\bf \textit{hyperbolas, vertical traverse axis }
\\\\
\cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1
\qquad 
\begin{cases}
center\ ( h, k)\\
vertices\ ( h,  k\pm a)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{ a ^2 + b ^2}\\
asymptotes\quad  y= k\pm \cfrac{a}{b}(x- h)
\end{cases}\\\\
-------------------------------

\bf \begin{cases}
h=2\\
k=-3\\
a=6\\
c=2\sqrt{10}
\end{cases}\implies \cfrac{[y- (-3)]^2}{ 6^2}-\cfrac{(x- 2)^2}{ b^2}=1
\\\\\\
\cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ b^2}=1
\\\\\\
c^2=a^2+b^2\implies (2\sqrt{10})^2=6^2+b^2\implies 2^2(\sqrt{10})^2=36+b^2
\\\\\\
4(10)=36+b^2\implies 40=36+b^2\implies 4=b^2
\\\\\\
\sqrt{4}=b\implies 2=b\\\\
-------------------------------\\\\
\cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 2^2}=1\implies \cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 4}=1

3 0
3 years ago
What is the difference between integers and real numbers?
galina1969 [7]
Integers are positive and negative whole numbers no fractions or decimals.
Real numbers do have fractions and decimals that integers do not.
5 0
3 years ago
Solve the system y = -x + 7 and y = -0.5(x - 3)^2 + 8.
AlladinOne [14]

Answer:

(1,6) & (7,0)

Step-by-step explanation:

y = -x + 7

y = -0.5(x - 3)² + 8

To solve the system, solve these two equations simultaneously

-x + 7 = -0.5(x - 3)² + 8

-x + 7 = -0.5(x² - 6x + 9) + 8

-x + 7 = -0.5x² + 3x - 4.5 + 8

0.5x² - 4x + 3.5 = 0

x² - 8x + 7 = 0

x² - 7x - x + 7 = 0

x(x - 7) - (x - 7) = 0

(x - 1)(x - 7) = 0

x = 1, 7

y = -1 + 7 = 6

y = -7 + 7 = 0

(1,6) (7,0)

Since the system has two distinct solutions, the line and the curve meet at two distinct poibts9: (1,6) & (7,0)

8 0
3 years ago
I need help with with problem please thanks!
liraira [26]
40 is the correct answer. Hope this helps :)
5 0
4 years ago
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