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3 years ago

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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A 10 m. ramp at an amusement park rises 2 m. in the air. What is the angle of elevation of the ramp?

larisa86 [58]

16 acute angle is the answer

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3 years ago

ILL GIVE BRAINLEST

iris [78.8K]

Answer: say "theres no graph" is you are a caniball

Step-by-step explanation:

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3 years ago

The formula

cricket20 [7]

Answer:

See below ~

Step-by-step explanation:

Solving for F

5/8 (F - 32) + 273.15 = K
5/8 (F - 32) = K - 273.15
F - 32 = 8/5 (K - 273.15)
F = 8/5 (K - 273.15) + 32

Finding F when K = 310.15 K

F = 8/5 (310.15 - 273.15) + 32
F = 1.6 (37) + 32
F = 59.2 + 32
F = 81.2°F

6 0

2 years ago

Please help I do not understand this at all and I need help

KengaRu [80]

Answer:

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

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

$\bf \textit{vertex of a parabola}\\ \quad \\

\begin{array}{lccclll}
h(t)=&-16t^2&+64t&+12\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\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\\\\
\begin{cases}
v_o=\textit{initial velocity of the object}\to &128\\
h_o=\textit{initial height of the object}\to &12\\
h=\textit{height of the object at "t" seconds}\end{cases}\\\\
-----------------------------\\\\
\textit{vertex of a parabola}\\ \quad \\

\begin{array}{lccclll}
h(t)=&-16t^2&+128t&+12\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\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