617513.4668 is the answer
Answer:
2125 miles
Step-by-step explanation:
Okay so to work this out we first need to find out how far the plane flies in one minute. To do this we do:
175 = 7
divide both sides by 7 to get 1 minute
25 = 1
The plane travels 25 miles in one minute.
Now to get 85 minutes we just do 25 x 85
25 x 85 = 2125
The plane travels 2125 miles in 85 minutes.
Hope this helps you :)
<span>place vertex at (0,0)
basic equation: x^2=4py
p=distance from vertex to focus)
y=at ends of parabolic cross section
y = x^2/4p
Given y= 1/48x^2
So,
1/48x^2=X*2/4p
1/4p = 1/48
4p = 48
p = 48/4 = 12 ft</span>
<u>Step-by-step explanation:</u>
transform the parent graph of f(x) = ln x into f(x) = - ln (x - 4) by shifting the parent graph 4 units to the right and reflecting over the x-axis
(???, 0): 0 = - ln (x - 4)
0 = ln (x - 4)
1 = x - 4
<u> +4 </u> <u> +4 </u>
5 = x
(5, 0)
(???, 1): 1 = - ln (x - 4)
1 = ln (x - 4)
e = x - 4
<u> +4 </u> <u> +4 </u>
e + 4 = x
6.72 = x
(6.72, 1)
Domain: x - 4 > 0
<u> +4 </u> <u>+4 </u>
x > 4
(4, ∞)
Vertical asymptotes: there are no vertical asymptotes for the parent function and the transformation did not alter that
No vertical asymptotes
*************************************************************************
transform the parent graph of f(x) = 3ˣ into f(x) = - 3ˣ⁺⁵ by shifting the parent graph 5 units to the left and reflecting over the x-axis
Domain: there is no restriction on x so domain is all real number
(-∞, ∞)
Range: there is a horizontal asymptote for the parent graph of y = 0 with range of y > 0. the transformation is a reflection over the x-axis so the horizontal asymptote is the same (y = 0) but the range changed to y < 0.
(-∞, 0)
Y-intercept is when x = 0:
f(x) = - 3ˣ⁺⁵
= - 3⁰⁺⁵
= - 3⁵
= -243
Horizontal Asymptote: y = 0 <em>(explanation above)</em>
Answer:
sin(2x)=cos(π2−2x)
So:
cos(π2−2x)=cos(3x)
Now we know that cos(x)=cos(±x) because cosine is an even function. So we see that
(π2−2x)=±3x
i)
π2=5x
x=π10
ii)
π2=−x
x=−π2
Similarly, sin(2x)=sin(2x−2π)=cos(π2−2x−2π)
So we see that
(π2−2x−2π)=±3x
iii)
π2−2π=5x
x=−310π
iv)
π2−2π=−x
x=2π−π2=32π
Finally, we note that the solutions must repeat every 2π because the original functions each repeat every 2π. (The sine function has period π so it has completed exactly two periods over an interval of length 2π. The cosine has period 23π so it has completed exactly three periods over an interval of length 2π. Hence, both functions repeat every 2π2π2π so every solution will repeat every 2π.)
So we get ∀n∈N
i) x=π10+2πn
ii) x=−π2+2πn
iii) x=−310π+2πn
(Note that solution (iv) is redundant since 32π+2πn=−π2+2π(n+1).)
So we conclude that there are really three solutions and then the periodic extensions of those three solutions.
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