Suppose that equation of parabola is
y =ax² + bx + c
Since parabola passes through the point (2,−15) then
−15 = 4a + 2b + c
Since parabola passes through the point (-5,-29), then
−29 = 25a − 5b + c
Since parabola passes through the point (−3,−5), then
−5 = 9a − 3b + c
Thus, we obtained following system:
4a + 2b + c = −15
25a − 5b + c = −29
9a − 3b + c = −5
Solving it we get that
a = −2, b = −4, c = 1
Thus, equation of parabola is
y = −2x²− 4x + 1
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Rewriting in the form of
(x - h)² = 4p(y - k)
i) -2x² - 4x + 1 = y
ii) -3x² - 7x = y - 11
(-3x² and -7x are isolated)
iii) -3x² - 7x - 49/36 = y - 1 - 49/36
(Adding -49/36 to both sides to get perfect square on LHS)
iv) -3(x² + 7/3x + 49/36) = y - 3
(Taking out -3 common from LHS)
v) -3(x + 7/6)² = y - 445/36
vi) (x + 7/6)² = -⅓(y - 445/36)
(Shifting -⅓ to RHS)
vii) (x + 1)² = 4(-1/12)(y - 445/36)
(Rewriting in the form of 4(-1/12) ; This is 4p)
So, after rewriting the equation would be -
(x + 7/6)² = 4(-⅛)(y - 445/36)
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I hope this is what you wanted.
Regards,
Divyanka♪
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Answer:
C
Step-by-step explanation:
Hope dis helps
Answer:
0
Step-by-step explanation:
anything times 0 is 0
Based on the graph given, the option that will show the same amplitude as function m is graph D.
<h3>Which graphed function is this about?</h3>
The cosine function is seen as:
f(x) = A*cos(kx) + M
And the functions are:
- A stands for amplitude,
- k is angular frequency,
- M is the midline.
When the function is m(x) = -2*cos(x+π).
The absolute value of the amplitude will be 2*|-2| = 4
Therefore, the option that can have the requirement above is graph D.
Learn more about cosine from
brainly.com/question/23563998
#SPJ4
a.

is a proper joint density function if, over its support,
is non-negative and the integral of
is 1. The first condition is easily met as long as
. To meet the second condition, we require

b. Find the marginal joint density of
and
by integrating the joint density with respect to
:


Then


c. This probability can be found by simply integrating the joint density:

