Answer:
Step-by-step explanation:
The system would be:
And I attached the graph. The system will have 2 solution (The 2 intercepts between the curves)
Stretched by 3, right 2, up 1
Let Fn be the number of ways of arranging such flagpole with the given conditions.
When arranging a flagpole of n feet high, consider the following cases
If the last flag used is a red flag, then the other flags are n-1 foot high, so they can be seen as arranged on a smaller flagpole of n-1 feet high, which can be done in Fn-1 ways.
Similarly, If the last flag used is a gold flag, then the other flags can be seen as arranged on a smaller flagpole of n-1 feet high. This can be done in Fn-1 ways.
If the last flag used is green, the other flags are n-2 feet high, so the flagpole can be arranged in Fn-2 ways.
Using the sum rule, we obtain that Fn = 2Fn-1 + Fn-2 for all n≥3. Listing all the combinations of flags, the initial conditions are F1 = 2 and F2 = 3.
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When two secant lines intersect each other outside a circle, the products of their segments are equal.
14(с + 14) = 10(10 + 20)
14с + 196 = 10 * 30
14с +196 = 300
14с = 300 - 196
14с = 104
с = 104/14
с ≈ 7.4