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3 years ago

How do i solve it pleas don't give the answer but give hints

Mathematics

1 answer:

Marta_Voda [28]3 years ago

6 0

Step-by-step explanation:

A mixed number consists of an integer and a proper fraction:

$a\frac{b}{c}$

where the numerator is less than the denominator (b < c).

A terminating decimal is a decimal that ends. For example, 1/2 = 0.5.

A repeating decimal is a decimal that repeats forever. For example, 1/3 = 0.33333...

You must use 2, 3, and 4 exactly once. Try different combinations and see which ones result in a terminating decimal, and which one results in a repeating decimal.

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Hi,

x^5 = x * x * x * x * x.

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Help meee pleasee!!!<br><br><br><br><br> thank you <3

julsineya [31]

The equation of the newsletter function is C(x) = 75 + 0.25x and the function values are C(0) = 75, C(100) = 100, C(200) = 125 and C(300) = 150

<h3 /><h3>How to determine the newsletter function?</h3>

From the question, the given parameters are

Initial charge = $75.00

Rate per copy = $0.25 per copy

The equation of the newsletter function is then calculated as

Total = Initial charge + Rate per copy x Number of copies

Let x represents the number of copies

So, we have

Total = Initial charge + Rate per copy x x

This gives

C(x) = 75 + 0.25x

<h3>The function values for x = 0, 100, 200 and 300</h3>

When x = 0, we have

C(0) = 75 + 0.25 x 0 = 75

When x = 100, we have

C(100) = 75 + 0.25 x 100 = 100

When x = 200, we have

C(200) = 75 + 0.25 x 200 = 125

When x = 300, we have

C(300) = 75 + 0.25 x 300 = 150

Read more about linear equations at

brainly.com/question/4074386

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1 year ago

The recursive function LaTeX: f\left(0\right)=1,\:f\left(n\right)=f\left(n-1\right)+2nf ( 0 ) = 1 , f ( n ) = f ( n − 1 ) + 2 n

m_a_m_a [10]

Question:

The recursive function $f(0) = 1$ , $f(n) = f(n-1) + 2n$ represents the nth term of a sequence. Determine the explicit function

Answer:

$f(n) = n^2 + n+1$

Step-by-step explanation:

Given

$f(0) = 1$

$f(n) = f(n-1) + 2n$

Required

Write an explicit formula

Let n = 1

$f(1) = f(1-1) + 2*1$

$f(1) = f(0) + 2$

$f(1) = 1 + 2 = 3$

Let n = 2

$f(2) = f(2-1) + 2*2$

$f(2) = f(1) + 4$

$f(2) = 3 + 4 = 7$

Let n =3

$f(3)=f(3-1) + 2 * 3$

$f(3)=f(2) + 6$

$f(3)=7+ 6 = 13$

Let n = 4

$f(4) = f(4 - 1) + 2 * 4$

$f(4) = f(3) + 8$

$f(4) = 13+ 8 = 21$

So, we have:

$f(1) = 3= 1 + 2 * 1=1+0*1*2*1$

$f(2) = 7= 3 + 2 * 2=1+1*2+2*2$

$f(3) = 13= 7 + 2 * 3= 1+2*3+2*3$

$f(4) = 21 = 13 + 2 * 4= 1+3*4 +2 *4$

Following the above pattern:

$f(n) = 1 + (n - 1) * n + 2 * n$

$f(n) = 1 + n(n - 1) + 2n$

Open bracket

$f(n) = 1 + n^2 - n + 2n$

$f(n) = 1 + n^2 + n$

$f(n) = n^2 + n+1$

3 0

2 years ago