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s2008m [1.1K]
4 years ago
15

За^n(а^n + a^n-1) what is it??

Mathematics
1 answer:
solniwko [45]4 years ago
8 0
Distribute the 3a^n to all the other values then solve...
(3a^n*a^n)+(3a^n*a^n)+(3a^n*-1)
4a^2n+4a^2n-3a^n=
8a^2n-3a^n
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Hey, the question is in the pic.
nikklg [1K]
The answer would be A

7 0
3 years ago
2 The diagram shows a
Strike441 [17]

Diagram? There is no diagram given.

a.

i =  \frac{180(n - 2)}{n}

i =  \frac{180(12 - 2)}{12}  =  \frac{180(10)}{12}  =  \frac{1800}{12}  \\ i = 150\degree

Cannot solve b without the diagram.

5 0
3 years ago
XYZ Toys makes a Feed-Me Baby Doll for $10.40. They mark up the cost by 55%. What is the selling price for the Feed-Me Baby Doll
Lesechka [4]
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3 0
3 years ago
A rock is thrown upward from a bridge that is 22 feet above a road. The rock reaches its maximum height above the road 0.69 seco
Flura [38]

Answer:

  • <u>f(t) = -7.561(t + 1.15)(t - 2.53)</u>

Explanation:

The<em> hint </em>is that the function f can be written in the form:

  • f(t) = c (t - t₁) (t - t₂)

It is a quadratic function, i.e. a parabola.

The maximum point, maximum height, of a parabola is its vertex.

And the x-coordinate of the vertex is at the midpoint between the two x-intercepts.

The time when the rock reaches the road is one x-intercept: t = 2.53 seconds: (2.53, 0).

The other x-intercept is sucht that (x + 2.53) / 2 = 0.69

  • (x + 2.53) / 2 = 0.69
  • x + 2.53 = 1.38
  • x = 1.38 - 2.53
  • x = -1.15

Thus, the equation is:

  • f(t) = c(t + 1.15) (t - 2.53).

You can find c by doing t = 0, because at t = 0 the height, f(t), is 22 feet:

  • f(0) = 22 = c (0 +1.15)(0 - 2.53)
  • 22 = c(1.15)(-2.53)
  • 22 = c (-2.9095)
  • c = - 22/(2.9095)
  • c = - 7.561

Then, the function can be written as:

  • f(t) = -7.561(t + 1.15) (t - 2.53) ← answer
6 0
3 years ago
N=4; 2i and 3i are zeros <br> f(-1)=50
Nonamiya [84]

Solution:- \text{Let f(x) be any nth degree polynomial with n=4}\\

\text{Given that 2i and 3i are the zeroes of f(x)}

\text{so (x-2i) and (x-3i) are factors of f(x)}

\text{Since 2i is a zero of f(x) then its conjugate -2i is also a zero of f(x)}

\Rightarrow(x+2i) \text{is a factor}\\\text{Similarly, conjugate of 3i is -3i is also a zero of f(x) }\\\Rightarrow(x+3i)\text{is a factor}\\\text{So , }\\f(x)=k(x-2i)(x+2i)(x-3i)(x+3i)\\=k(x^2-(2i)^2)(x^2-(3i)^2)\\=k(x^2-4i^2)(x^2-9i^2)\\=k(x^2+4)(x^2+9)\\=k(x^4+13x^2+36)\\\text{As given}\\f(-1)=50\\\Rightarrow k((-1)^4+13(-1)^2+36)=50\\\Rightarrow k(1+13+36)=50\\\Rightarrow k(50)=50\\\Rightarrow k=1\\\text{So by substituting k=1 in f(x) we get ,}\\f(x)=(x^4+13x^2+36)

6 0
3 years ago
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