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4 years ago

41 9/11 minus 27 1/3

Mathematics

2 answers:

sertanlavr [38]4 years ago

7 0

41 and 9/11 - 27 1/3 is equal to 14and 6/33

Hope it helps!

Think about what I gave you!:D

GrogVix [38]4 years ago

3 0

The answer is 20 16/33 or in improper form it is 676/33.

You might be interested in

function in c that takes three decimal numbers as inputs and finds the sum of those inputs and returns the sum

kozerog [31]

The function that takes three decimal numbers as inputs and finds the sum of those inputs and returns the sum has been given below

What is programming?

Program is the set of instructions given to the computer so that the computer can execute those instructions. The act of writing program is called programming.

The program has been shown below

#include<stdio.h>

using namespace std;

int main()

{

double n1, n2, n3;

double s;

printf("Enter first numbers");

scanf(%lf, &n1);

printf("Enter second numbers");

scanf(%lf, &n2);

printf("Enter third numbers");

scanf(%lf, &n3);

s = n1 + n2 + n3;

printf("Sum = %f", s);

return 0;

}

To learn more about programming refer to the link-

brainly.com/question/23275071

#SPJ4

7 0

1 year ago

The base of a parallelogram is 10 inches the height is 2 inches more than half the base find the area

Damm [24]

Given is a parallelogram whose base is b = 10 inches and it says height is 2 more than half-value of base i.e. h = 2 + (b ÷ 2).

So height is h = 2 + (10 ÷ 2) = 2 + 5

⇒ h = 7 inches.

We know the formula for area of parallelogram is given as follows :-

Area of parallelogram = base x height

Area of parallelogram = 10 inches x 7 inches

Area of parallelogram = 70 squared inches.

Hence, the final answer is 70 squared inches.

8 0

3 years ago

Ashton purchased a new vacuum for $1,150 at the Best Buy. He has a 15% coupon. How much will he save?

Kipish [7]

Answer:

$172.50

Step-by-step explanation:

1150(0.15)

172.50

6 0

3 years ago

At the state fair, admission at the gate is $9. In addition, the cost of each ride is $3. Suppose that Deshaun will go on x ride

pshichka [43]

SOLUTION:

Step 1:

In this question, we are given the following:

Step 2:

The details of the solution are as follows:

Step-by-step explanation:

Given :

At the state fair, admission at the gate is $9.

In addition, the cost of each ride is $3.

Suppose that Sam will go on x rides.

Then, cost of x rides ( in dollars) = Cost per ride x Number of rides

=3x

Total cost ( in dollars) : Admission fee + cost of x rides

= 9+3x

Sam wants the total number of dollars he spends on admission and rides to be at most t ( less than or equal to t )

$9\text{ + 3x }\leq\text{ t}$

CONCLUSION:

The final answer is:

Using the values and variables given, write an inequality describing this.

$9\text{ + 3x }\leq\text{ t}$

4 0

1 year ago

For the function defined by f(t)=2-t, 0≤t<1, sketch 3 periods and find:

Oksi-84 [34.3K]

The half-range sine series is the expansion for $f(t)$ with the assumption that $f(t)$ is considered to be an odd function over its full range, $-1$ . So for (a), you're essentially finding the full range expansion of the function

$f(t)=\begin{cases}2-t&\text{for }0\le t$

with period 2 so that $f(t)=f(t+2n)$ for $|t|$ and integers $n$ .

Now, since $f(t)$ is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with

$f(t)=\displaystyle\sum_{n\ge1}b_n\sin\frac{n\pi t}L$

where

$b_n=\displaystyle\frac2L\int_0^Lf(t)\sin\frac{n\pi t}L\,\mathrm dt$

In this case, $L=1$ , so

$b_n=\displaystyle2\int_0^1(2-t)\sin n\pi t\,\mathrm dt$
$b_n=\dfrac4{n\pi}-\dfrac{2\cos n\pi}{n\pi}-\dfrac{2\sin n\pi}{n^2\pi^2}$
$b_n=\dfrac{4-2(-1)^n}{n\pi}$

The half-range sine series expansion for $f(t)$ is then

$f(t)\sim\displaystyle\sum_{n\ge1}\frac{4-2(-1)^n}{n\pi}\sin n\pi t$

which can be further simplified by considering the even/odd cases of $n$ , but there's no need for that here.

The half-range cosine series is computed similarly, this time assuming $f(t)$ is even/symmetric across its full range. In other words, you are finding the full range series expansion for

$f(t)=\begin{cases}2-t&\text{for }0\le t$

Now the sine series expansion vanishes, leaving you with

$f(t)\sim\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi t}L$

where

$a_n=\displaystyle\frac2L\int_0^Lf(t)\cos\frac{n\pi t}L\,\mathrm dt$

for $n\ge0$ . Again, $L=1$ . You should find that

$a_0=\displaystyle2\int_0^1(2-t)\,\mathrm dt=3$

$a_n=\displaystyle2\int_0^1(2-t)\cos n\pi t\,\mathrm dt$
$a_n=\dfrac2{n^2\pi^2}-\dfrac{2\cos n\pi}{n^2\pi^2}+\dfrac{2\sin n\pi}{n\pi}$
$a_n=\dfrac{2-2(-1)^n}{n^2\pi^2}$

Here, splitting into even/odd cases actually reduces this further. Notice that when $n$ is even, the expression above simplifies to

$a_{n=2k}=\dfrac{2-2(-1)^{2k}}{(2k)^2\pi^2}=0$

while for odd $n$ , you have

$a_{n=2k-1}=\dfrac{2-2(-1)^{2k-1}}{(2k-1)^2\pi^2}=\dfrac4{(2k-1)^2\pi^2}$

So the half-range cosine series expansion would be

$f(t)\sim\dfrac32+\displaystyle\sum_{n\ge1}a_n\cos n\pi t$
$f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}a_{2k-1}\cos(2k-1)\pi t$
$f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}\frac4{(2k-1)^2\pi^2}\cos(2k-1)\pi t$

Attached are plots of the first few terms of each series overlaid onto plots of $f(t)$ . In the half-range sine series (right), I use $n=10$ terms, and in the half-range cosine series (left), I use $k=2$ or $n=2(2)-1=3$ terms. (It's a bit more difficult to distinguish $f(t)$ from the latter because the cosine series converges so much faster.)

5 0

4 years ago