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3 years ago

Help me with 32 please!!!!!!

Mathematics

1 answer:

Alexeev081 [22]3 years ago

8 0

-8x - 46 = -6x - 8
-8x + 6x = -8 + 46
-2x = 38
x = -19

The equation:
-8x - 46 = -6x - 8

The solution:
x = -19

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Billy had 55 stamps in his collection. Billy’s older brother Tommy had 73 stamps in his collection. How many stamps did the two

KengaRu [80]

Answer:

128

Step-by-step explanation:

Billy has 55 stamps and his brother Tommy has 73. Add 73 and 55 together. 55+73=128

6 0

1 year ago

A certain type of brass contains 65% copper. How many pounds of copper are contained in 120 pounds of brass?

enyata [817]

The number of pounds of copper in 120 pounds of brass is = 78 pounds of copper

<h3>Calculation of number of pounds of copper</h3>

The percentage of copper in the brass = 65%

Therefore the quantity of copper in the brass = 65/100= 0.65 pounds

This means that 1 pound brass contains = 0.65 pounds of copper

The quantity of copper in 120 pounds of brass = 120 * 0.65

= 78 pounds of copper.

Learn more about copper here:

brainly.com/question/24856041

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7 0

1 year ago

Help on # three it’s due in 10 minuets!!

LuckyWell [14K]

16 = 48/3
48 divided by 2 = 24
There is 24 students in the club.

6 0

3 years ago

Can I get help with finding the Fourier cosine series of F(x) = x - x^2

trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

$f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx$

You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
$a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}$
$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
$a_0=-\dfrac{2L^2}3$

$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

4 0

2 years ago

Please answer the question, and explain each step

natta225 [31]

The steps are shown here.

1) Factoring
2) Completing the square
3) Quadratic Formula
4) Graphing

So basically the answer would either be 5 or -5.

5 0

2 years ago