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kaheart [24]
3 years ago
15

What constant term should be used to complete the square? x 2 - 5x + _____ = 7

Mathematics
1 answer:
amm18123 years ago
4 0
The quadratic equation in its generic form is:
 ax2 + bx + c
 To complete squares we must add the following term:
 (b / 2) ^ 2
 The equation is:
 ax2 + bx + c + (b / 2) ^ 2
 We have the following equation:
 x ^ 2 - 5x + k = 7
 By completing squares we have:
 x ^ 2 - 5x + (-5/2) ^ 2 = 7 + (-5/2) ^ 2
 Rewriting:
 x ^ 2 - 5x + 6.25 = 7 + 6.25
 Answer:
 
A constant term should be used to complete the square is:
 
6.25
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Answer:

$3

Step-by-step explanation:

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To solve, multiply 20/100 (20%) by 15:

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JulijaS [17]

Answer:

c

Step-by-step explanation:

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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
Write an equation of the line containing the given point and perpendicular to the given line.
Tresset [83]

Answer:

Step-by-step explanation:

The equation of a straight line can be represented in the slope intercept form as

y = mx + c

Where

m = slope = (change in the value of y on the vertical axis) / (change in the value of x on the horizontal axis)

The equation of the given line is

9x+7y=4

7y = 4 - 9x = -9x + 4

y = -9x/7 + 4/7

Comparing with the slope intercept form, slope = -9/7

If the line passing through the given point is perpendicular to the given line, it means that its slope is the negative reciprocal of the slope of the given line.

Therefore, the slope of the line passing through (7,-4) is 7/9

To determine the intercept, we would substitute m = 7/9, x = 7 and y = - 4 into y = mx + c. It becomes

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c = - 4 - 49/9

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The equation becomes

y = 7x/9 - 85/9

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JulijaS [17]

Answer:

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Step-by-step explanation:

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