All categories

1 year ago

Paula has buttons. she gives away 28 buttons. choose the expression that shows the number of buttons paula has left.

Mathematics

1 answer:

mamaluj [8]1 year ago

3 0

Answer:

y = 28 - x

Step-by-step explanation:

The variable x represents how many buttons Paula had before giving 28 away and the variable y represents how many buttons she has now.

You might be interested in

A six-sided number cube is rolled and a coin is tossed. What is the probability getting a 6 and a heads?

Whitepunk [10]

Cube= 1/6
coin= 1/2
1/2*3= 3/6
3/6+1/6= 4/6
4/6= 2/3
The answer is 4/6 or 2/3

6 0

2 years ago

Read 2 more answers

Help me fast times-a-ticking

d1i1m1o1n [39]

No, because it has a constant rate of change

By looking at this table

The y value changes at a constant multiple of 2

6 4

7 2 (4-2 is a difference of 2)

8 0 (2-0 is a difference of 2)

9 -2 (0-2 is a difference of 2)

This means it is Linear

8 0

3 years ago

Read 2 more answers

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

3 0

2 years ago

NEED TO SUBMIT NOW PLS HELPPP

kherson [118]

Answer: 40

Step-by-step explanation:

138/x = 207/60

8280 = 207x

divide by 207

40 = x

7 0

2 years ago

The length of each edge of a cube is x centimeters. If x is an integer, why is the volume of the cube not equal to 15m^3?

Nadusha1986 [10]

Suppose the length of each side of the cube is x, so the volume must be:

x^3.

If the volume is 15 cm^3, so the length must be cuberoot(15) which is not integer,. So, the volume of the cube with integer side never equal with 15 cm^3

4 0

2 years ago