The statement is false because when you use distributive property, you get -12=-4+16 and when you add 16 to -4, you'll get 12, not -12, making the statement false.
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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Alright, so for AB and CD to be parallel, CX and DX would have to be equal, as is with AX and BX. In addition, for CD and AB to be parallel, all sides in both triangles are either equal or not all sides in even one triangle are equal. Therefore, CD is not 3. In addition, two sides of a triangle combined must be greater than the third, so that leaves 5, 4, and 2 for CD. If it was 5, that would mean that all sides are equal, so that leaves 4 and 2. However, I don't see anything to prove either one right, sorry:/
Answer:
3 : 5
Step-by-step explanation:
The ratio is 9 : 15 , but since we're simplifying to lowest terms, I divided each value by 3
