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antiseptic1488 [7]
3 years ago
11

Can someone help idk what this is

Mathematics
1 answer:
Anarel [89]3 years ago
7 0

Answer:

x=1 y=12

Step-by-step explanation:

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.Given the functions f(n) = 500 and g(n) = (nine tenths)n _ 1, combine them to create a geometric sequence, an, and solve for th
weeeeeb [17]
The original functions are: f(n) = 500 and g(n) = [9/10]^(n-1)

A geometric sequence combining them is: An = f(n)*g(n) = 500*[9/10]^(n-1):

Some terms are:
A1= 500
A2 = 500*[9/10]
A3 = 500*[9/10]^2
A4 = 500*[9/10]^3
....
A11 = 500*[9/10]^10 ≈ 174.339

Answer: the third option, An = 500[9/10]^(n-1); A11 = 174.339

 
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Figure ABCDE is similar to figure FGHIJ. What is the length of segment FG?
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Which statement best describes the function represented by the graph??
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The last one decreasing on the interval (-infinity, 0) where it becomes constant

Step-by-step explanation:

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15 ÷ 3 = Δ
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this is the answer

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Let f(x)=x^3-x-1
alexira [117]
f(x) = x^3 - x - 1

To find the gradient of the tangent, we must first differentiate the function.

f'(x) = \frac{d}{dx}(x^3 - x - 1) = 3x^2 - 1

The gradient at x = 0 is given by evaluating f'(0).

f'(0) = 3(0)^2 - 1 = -1

The derivative of the function at this point is negative, which tells us <em>the function is decreasing at that point</em>.

The tangent to the line is a straight line, so we will have a linear equation of the form y = mx + c. We know the gradient, m, is equal to -1, so

y = -x + c

Now we need to substitute a point on the tangent into this equation to find c. We know a point when x = 0 lies on here. To find the y-coordinate of this point we need to evaluate f(0).

f(0) = (0)^3 - (0) - 1 = -1

So the point (0, -1) lies on the tangent. Substituting into the tangent equation:

y = -x + c \\\\ -1 = -(0) + c \\\\ -1 = c \\\\ \text{Equation of tangent is } \boxed{y = -x - 1}
6 0
3 years ago
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