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Rasek [7]
3 years ago
14

I think I know how to solve this, but could someone please check my work for me?

Mathematics
1 answer:
Fiesta28 [93]3 years ago
5 0
Answer is :

 cos⁻¹[(-1-4√7)/(√17, √8) = 173.33° ≈ 173°

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f(x)=4x+1\\\\g(x)=x^2-5\\\\(g/f)(x)=\dfrac{g(x)}{f(x)}=\dfrac{x^2-5}{4x+1}

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360

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The parabola y=x2y=x^2 y=x 2 y, equals, x, squared is scaled vertically by a factor of 110\dfrac{1}{10} 10 1 ​ start fraction, 1
sweet [91]

Question is not well presented

The parabola y=x² is scaled vertically by a factor of 1/10.

What is the equation of the new parabola?

Answer:

The equation of the new parabola is 0.1x²

Step-by-step explanation:

Given

Parabola: y = x²

Scale = 1/10 = 0.1

The interpretation of this question is that; there's a need to scale the graph in ratio 1:10.

I.e; 1 unit on the parabola is being represented by 10 unit on the scale

So, x (on the new parabola) = 1/10 of old x.

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3 years ago
The sequence {an) is defined by ao = 1 and
otez555 [7]
<h3>Answer: 31</h3>

=======================================

Work Shown:

Use the value of a0 to find the value of a1. The idea is you double the previous value, and then add 1.

a_{n+1} = 2*(a_n) + 1\\\\a_{0+1} = 2*(a_0) + 1\\\\a_{1} = 2*(1) + 1\\\\a_{1} = 3\\\\

Which is then used to find the value of a2. Follow the same process as before (double the previous value and then add 1).

a_{n+1} = 2*(a_n) + 1\\\\a_{1+1} = 2*(a_1) + 1\\\\a_{2} = 2*(3) + 1\\\\a_{2} = 7\\\\

This is used to find a3

a_{n+1} = 2*(a_n) + 1\\\\a_{2+1} = 2*(a_2) + 1\\\\a_{3} = 2*(7) + 1\\\\a_{3} = 15\\\\

Finally we can now find a4

a_{n+1} = 2*(a_n) + 1\\\\a_{3+1} = 2*(a_3) + 1\\\\a_{4} = 2*(15) + 1\\\\a_{4} = 31\\\\

Recursive sequences like this aren't too bad if n is small, but as n gets larger, things become more tedious. For those cases, its best to try to find a closed form equation. If not, then the next best thing is using a spreadsheet to automate the process.

6 0
3 years ago
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