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erastova [34]
3 years ago
7

Find the 100th term in the arithmetic sequence -19,-14,-9,-4,1...

Mathematics
1 answer:
Alex73 [517]3 years ago
4 0

Answer:

476

Step-by-step explanation:

hello :

the common diference is : d = ( 1)-(-4)=(-4)-(-9)=(-9)-(-14)=(-14)-(-19)=5

the firt term is A1 = -19

the n-ieme term is : An = A1 +(n-1)d

An = -19+5(n-1)

An = 5n-24

the 100th term in the arithmetic sequence  is : A100 = 5(100)-24 =476

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The function f(x) = x2 - 6x + 9 is shifted 5 units to the left to create g(x). What is g(x)?
creativ13 [48]

Answer:

g(x) = x^2 + 4x + 4

Step-by-step explanation:

In translation of functions, adding a constant to the domain values (x) of a function will move the graph to the left, while subtracting from the input of the function will move the graph to the right.

Given the function;

f(x) = x2 - 6x + 9

a shift 5 units to the left implies that we shall be adding the constant 5 to the x values of the function;

g(x) = f(x+5)

g(x) = (x+5)^2 - 6(x+5) + 9

g(x) = x^2 + 10x + 25 - 6x -30 + 9

g(x) = x^2 + 4x + 4

3 0
3 years ago
If you know the perimeter of the square, you can find the area of the square in two steps. Describe the two steps.
Veronika [31]
1) Take the perimeter and divide it by 4.
2) Take the answer from step 1 and square it (multiply by itself)
6 0
3 years ago
Read 2 more answers
Please help!! 20 POINTS
rjkz [21]

f(-3) would be 36.

When looking at synthetic division, the numbers across the top represent the coefficients of x^2, x and the constant in that order. Therefore, the equation is as follows.

2x^2 - 5x + 3

Now we can put -3 into the equation and solve.

2(-3)^2 - 5(-3) + 3

2(9) + 15 + 3

18 + 15 + 3

36

7 0
3 years ago
+
lesya692 [45]
The student would earn %86. To calculate percentage you take the amount of questions correct, divide that by the amount of questions given, multiply by 100 and round to the nearest percent.
3 0
3 years ago
Find maclaurin series
Mumz [18]

Recall the Maclaurin expansion for cos(x), valid for all real x :

\displaystyle \cos(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}

Then replacing x with √5 x (I'm assuming you mean √5 times x, and not √(5x)) gives

\displaystyle \cos\left(\sqrt 5\,x\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt5\,x\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^{2n}}{(2n)!}

The first 3 terms of the series are

\cos\left(\sqrt5\,x\right) \approx 1 - \dfrac{5x^2}2 + \dfrac{25x^4}{24}

and the general n-th term is as shown in the series.

In case you did mean cos(√(5x)), we would instead end up with

\displaystyle \cos\left(\sqrt{5x}\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt{5x}\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^n}{(2n)!}

which amounts to replacing the x with √x in the expansion of cos(√5 x) :

\cos\left(\sqrt{5x}\right) \approx 1 - \dfrac{5x}2 + \dfrac{25x^2}{24}

7 0
2 years ago
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