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Zarrin [17]
3 years ago
15

F(x)=-2x graph the parabola

Mathematics
1 answer:
andreev551 [17]3 years ago
8 0

Answer:

If core cm used et John week x en bc several

Step-by-step explanation:

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16 acute angle is the answer
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ILL GIVE BRAINLEST ​
iris [78.8K]

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Step-by-step explanation:

8 0
3 years ago
The formula
cricket20 [7]

Answer:

See below ~

Step-by-step explanation:

<u>Solving for F</u>

  • 5/8 (F - 32) + 273.15 = K
  • 5/8 (F - 32) = K - 273.15
  • F - 32 = 8/5 (K - 273.15)
  • <u>F = 8/5 (K - 273.15) + 32</u>

<u></u>

<u>Finding F when K = 310.15 K</u>

  • F = 8/5 (310.15 - 273.15) + 32
  • F = 1.6 (37) + 32
  • F = 59.2 + 32
  • F = <u>81.2°F</u>
6 0
2 years ago
Please help I do not understand this at all and I need help​
KengaRu [80]

Answer:

c

Step-by-step explanation:

28+35+21+12=96

7 0
3 years ago
How do I find the Velocity and How long will it take for the ball to reach it's maximum height?
Colt1911 [192]
So hmm check the picture below

\bf \qquad \textit{initial velocity}\\\\&#10;h = -16t^2+v_ot+h_o \qquad \text{in feet}\\&#10;\\ &#10;\begin{cases}&#10;v_o=\textit{initial velocity of the object}\to &64\\&#10;h_o=\textit{initial height of the object}\to &12\\&#10;h=\textit{height of the object at "t" seconds} \end{cases}\\\\&#10;-----------------------------\\\\

\bf \textit{vertex of a parabola}\\ \quad \\&#10;&#10;\begin{array}{lccclll}&#10;h(t)=&-16t^2&+64t&+12\\&#10;&\uparrow &\uparrow &\uparrow \\&#10;&a&b&c&#10;\end{array}\qquad &#10;\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad  {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)

part 1)  

it takes  \bf -\cfrac{{{ b}}}{2{{ a}}}\quad seconds

part 2)

\bf \textit{now, doubling }v_o\\\\&#10;\begin{cases}&#10;v_o=\textit{initial velocity of the object}\to &128\\&#10;h_o=\textit{initial height of the object}\to &12\\&#10;h=\textit{height of the object at "t" seconds}\end{cases}\\\\&#10;-----------------------------\\\\&#10;\textit{vertex of a parabola}\\ \quad \\&#10;&#10;\begin{array}{lccclll}&#10;h(t)=&-16t^2&+128t&+12\\&#10;&\uparrow &\uparrow &\uparrow \\&#10;&a&b&c&#10;\end{array}\qquad &#10;\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad  {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)

it will reach the maximum height at   \bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\quad feet


how much higher than before is that? well, what was the y-coordinate for when the vₒ was 64? what did you get for \bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}} ?

subtract that from this height when vₒ is 128 or doubled, to get their difference, that's how much higher it became

4 0
3 years ago
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