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:
x = 4.5
Step-by-step explanation:
14 = 8/3x + 2
12 = 8/3x
36 = 8x
x = 4.5
Answer: The maximum possible value of y is 30 ;)
Answer:
8 packs of bottled water and 14 packs of candy bars
Step-by-step explanation:
Provided,
Each pack of water bottles = 35 bottles
Each pack of candy = 20 candy
Since quantity of candy in each pack is small than quantity of water bottle in each pack, in order to have same quantity in numbers of both bottles and candies,
Purchase quantity of candy packs shall be more than the pack of water bottles.
Further, with this option 1 and 2 are not valid as candy packs are same or less than packs of water bottle.
In option 3 and 4, option 3 have smaller quantities as provided manager bought the least possible quantity.
Accordingly
Option 3
8 packs of water bottle = 
14 packs of candy = 
Since the quantities of water bottles and candies are same this is an opt answer.
<span>it all looks confusing when we try to juggle with all those numbers in the head. The problem can be solved systematically by constructing a contingency table.
</span>role/gender B G total
speaking...... 4 4 8
<span> silent............ 4 8 12
total............. 8 12 20
</span>Probability of a child having a speaking part is therefore
(4+4)/20=8/20=2/5
a. 2/5
