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Mariulka [41]
3 years ago
5

What equation models it’s position, y, with respect to time, x?

Mathematics
2 answers:
Nuetrik [128]3 years ago
5 0
The answer is the first one.
mr_godi [17]3 years ago
4 0

Answer:

i would say the first one

Step-by-step explanation:

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Given two right triangle legs
Use the Pythagorean theorem to calculate the hypotenuse from right triangle sides. Take a square root of sum of squares:
c = √(a² + b²)
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3 years ago
Solve the following two-step problems -5m - 4 = 4m
Musya8 [376]

Answer:

0.444

Step-by-step explanation:

3 0
2 years ago
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PLS ANSWER RN
Rzqust [24]

Answer:

So, the answer in short is B) (6,1)

Step-by-step explanation:

If you graph the points, then you see that point A is 8 points away from point B on the X-axis. If you take 8 and divide it by 4, you get 2, so that means that the 1/4 way point is 2 points away from point A, giving you 6 as your x-axis point for this new point. B is the only one with 6 as its spot on the x axis, leaving B as the only choice. BTW you're pretty cute ;)

6 0
3 years ago
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Help !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
murzikaleks [220]

Answer:

22.7

Step-by-step explanation:

Okay so the situation given shows a right triangle with the two sides being 7.6 and 18.2. So we can find the angle of elevation using tangent, which is tan-1*7.6/18.2. I got 22.66 so 22.7

3 0
2 years ago
Find the equation of a parabola that has a vertex of (-2,-3) and contains the point (4,1)
sweet [91]
\bf ~~~~~~\textit{parabola vertex form}
\\\\
\begin{array}{llll}
\boxed{y=a(x- h)^2+ k}\\\\
x=a(y- k)^2+ h
\end{array}
\qquad\qquad
vertex~~(\stackrel{-2}{ h},\stackrel{-3}{ k})\\\\
-------------------------------\\\\
\begin{cases}
h=-2\\
k=-3
\end{cases}\implies y=a[x-(-2)]^2-3\implies y=a(x+2)^2-3
\\\\\\
\textit{we also know that }
\begin{cases}
x=4\\
y=1
\end{cases}\implies 1=a(4+2)^2-3
\\\\\\
4=36a\implies \cfrac{4}{36}=a\implies \cfrac{1}{9}=a
\\\\\\
therefore\qquad \boxed{y=\cfrac{1}{9}(x+2)^2-3}
 
\bf y=\cfrac{1}{9}(x+2)^2-3\implies y=\cfrac{1}{9}(x^2+4x+4)-3
\\\\\\
y=\cfrac{1}{9}x^2+\cfrac{4}{9}x+\cfrac{4}{9}-3\implies \stackrel{standard~form}{y=\cfrac{1}{9}x^2+\cfrac{4}{9}x-\cfrac{23}{9}}
5 0
3 years ago
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