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Anuta_ua [19.1K]
3 years ago
15

For Thanksgiving, a family of 9 is trying to share 4 store-bought pies fairly. If each pie is pre-cut into 8 slices, and everyon

e gets the same number of slices, how many slices will be left over?
Mathematics
1 answer:
prisoha [69]3 years ago
3 0

Answer:

5

Step-by-step explanation:

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Maritza's car has 16 gallons of gas in the tank. She uses 3/4 of the gas. How many gallons of gas does Maritza use?
Savatey [412]
Find 3/4 of 16.

16 * 3/4

48/4

12

<span>So she uses 12 gallons of gas.</span>
5 0
3 years ago
Find the area of this circle please…Use 3 for pi
ZanzabumX [31]

Given:

The radius of circle is 11 inch

The value for pi is 3

The objective is to find the area of the circle.

The area of the circle is given by the formula:

A=\pi\times r^2^{}

Substituting ,r = 11

\pi=3

We get:

\begin{gathered} A=\pi\times r^2 \\ =3\times(11)^2 \\ =3\times121 \\ =363 \end{gathered}

Hence, the area of the circle is 363 square inches

5 0
1 year ago
DUE IN 10 MINUTES!! PLEASE ANSWER THESE QUESTIONS BELOW. 100 POINTS!!
expeople1 [14]

Answer:

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8 0
2 years ago
Find the half range Fourier sine series of the function
worty [1.4K]
The full range is -\pi (length 2L=2\pi), so the half range is L=\pi. The half range sine series would then be given by

f(x)=\displaystyle\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L=\sum_{n\ge1}b_n\sin nx

where

b_n=\displaystyle\frac2L\int_0^Lf(x)\sin\dfrac{n\pi x}L\,\mathrm dx=\frac2\pi\int_0^\pi(\pi-x)\sin nx\,\mathrm dx

Essentially, this is the same as finding the Fourier series for the function

\begin{cases}g(x)=\begin{cases}\pi-x&\text{for }0

Integrating by parts yields

b_n=\dfrac2\pi\left(\dfrac\pi n-\dfrac{\sin n\pi}{n^2}\right)=\dfrac2n

So the half range sine series for this function is simply

f(x)=\displaystyle\sum_{n\ge1}\frac{2\sin nx}n
5 0
3 years ago
The first three terms of an arithmetic series are 6p+2, 4p²-10 and 4p+3 respectively. Find the possible values of p. Calculate t
Doss [256]

Answer:

First Case:

\displaystyle p=\frac{5}{2}\text{ and } d=-2

Second Case:

\displaystyle p=-\frac{5}{4}\text{ and } d=\frac{7}{4}

Step-by-step explanation:

We know that the first three terms of an arithmetic series are:

6p+2, 4p^2-10, \text{ and } 4p+3

Since this is an arithmetic sequence, each subsequent term is <em>d</em> more than the previous term, where <em>d</em> is our common difference.

Therefore, we can write the second term as;

4p^2-10=(6p+2)+d

And, likewise, for the third term:

4p+3=(6p+2)+2d

Let's solve for <em>d</em> for each of the equations.

Subtracting in the first equation yields:

d=4p^2-6p-12

And for the second equation:

2d=-2p+1

To avoid fractions, let's multiply the first equation by 2. Hence:

2d=8p^2-12p-24

Therefore:

8p^2-12p-24=-2p+1

Simplifying yields:

8p^2-10p-25=0

Solve for <em>p</em>. We can factor:

8p^2+10p-20p-25=0

Factor:

2p(4p+5)-5(4p+5)=0

Grouping:

(2p-5)(4p+5)=0

Zero Product Property:

\displaystyle p_1=\frac{5}{2} \text{ or } p_2=-\frac{5}{4}

Then, we can use the second equation to solve for <em>d</em>. So:

2d_1=-2p_1+1

Substituting:

\begin{aligned} 2d_1&=-2(\frac{5}{2})+1 \\ 2d_1&=-5+1 \\ 2d_1&=-4 \\ d_1&=-2\end{aligned}

So, for the first case, <em>p</em> is 5/2 and <em>d</em> is -2.

Likewise, for the second case:

\begin{aligned} 2d_2&=-2(-\frac{5}{4})+1 \\ 2d_2&=\frac{5}{2}+1 \\ 2d_2&=\frac{7}{2} \\ d_2&=\frac{7}{4}\end{aligned}

So, for the second case, <em>p </em>is -5/4, and <em>d</em> is 7/4.

By using the values, we can determine our series.

For Case 1, we will have:

17, 15, 13.

For Case 2, we will have:

-11/2, -15/4, -2.

8 0
2 years ago
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