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3 years ago

14

there are 175 students at south tech high school. 100 students belong to neither the computer club or the car club. 55 students

are in the car club. how many students are only in the computer club?

1 answer:

Mumz [18]3 years ago

6 0

With only the description, my answer is 20 students.

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Sam takes x minutes yo walk from home to library and 5 minutes more to walk from the library to school than fron home to the lib

LuckyWell [14K]

Answer:

$2x+5\ minutes$

Step-by-step explanation:

Given:

Sam takes x minutes yo walk from home to library.

And takes 5 minutes more to walk from the library to school than from home to the library.

Express the total time taken for Sam to walk from home to school via the library in terms of x.

<u>Solution:</u>

Time taken by Sam to walk from home to library = $x\ minutes$

Time taken by Sam to walk from library to school = $x+5\ minutes$

Total time taken by Sam to walk from home to school via the library = Time taken to walk from home to library + Time taken to walk from library to school

Total time taken = $x+x+5=2x+5$

Therefore, the total time taken for Sam to walk from home to school via the library is $2x+5\ minutes$

8 0

2 years ago

Maya would like to invest her money in a savings account. The function A(t)=500(1.015)t represents the amount of money Maya will

Kobotan [32]

The answer is 6 years i know this because i made this problem

3 0

3 years ago

1/2 (12+8) • 52 - 17=_

qwelly [4]

The answer to your question is 503

5 0

2 years ago

Read 2 more answers

An ounce of milk contains 20 milligrams of calcium and 60 micrograms of vitamin A.

vova2212 [387]

Answer:

10 ounces of milk and 50 ounces of orange juice

Step-by-step explanation:

Let the amount of milk be x ounces and the amount of orange juice be y ounces.

1. An ounce of milk contains 20 milligrams of calcium and 60 micrograms of vitamin A. Then x ounces of milk contain 20x milligrams of calcium and 60x micrograms of vitamin A.

2. An ounce of orange juice contains 40 milligrams of calcium and 10 micrograms of vitamin A. Then y ounces of orange juice contain 40y milligrams of calcium and 10y micrograms of vitamin A.

<u>Calcium:</u>

20x milligrams in milk

40y milligrams in orange juice

2,200 milligrams in total

Hence,

$20x+40y=2,200$

<u>Vitamin A:</u>

60x micrograms in milk

10y micrograms in orange juice

1,100 micrograms in total

Hence,

$60x+10y=1,100$

Solve the system of two equations:

$\left\{\begin{array}{l}20x+40y=2,200\\ \\60x+10y=1,100\end{array}\right.$

Divide each equation by 10:

$\left\{\begin{array}{l}2x+4y=220\\ \\6x+y=110\end{array}\right.$

From the second equation:

$y=110-6x$

Substitute it into the first equation:

$2x+4(110-6x)=220\\ \\2x+440-24x=220\\ \\2x-24x=220-440\\ \\-22x=-220\\ \\22x=220\\ \\x=10\\ \\y=110-6\cdot 10=110-60=50$

6 0

3 years ago

How do I find the integral<br> ∫10(x−1)(x2+9)dxint10/((x-1)(x^2+9))dx ?

defon

$\int\frac{10}{(x-1)(x^2+9)}\ dx=(*)\\\\\frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}=\frac{A(x^2+9)+(Bx+C)(x-1)}{(x-1)(x^2+9)}\\\\=\frac{Ax^2+9A+Bx^2-Bx+Cx-C}{(x-1)(x^2+9)}=\frac{(A+B)x^2+(-B+C)x+(9A-C)}{(x-1)(x^2+9)}\\\Updownarrow\\10=(A+B)x^2+(-B+C)x+(9A-C)\\\Updownarrow\\A+B=0\ and\ -B+C=0\ and\ 9A-C=10\\A=-B\ and\ C=B\to9(-B)-B=10\to-10B=10\to B=-1\\A=-(-1)=1\ and\ C=-1$

$(*)=\int\left(\frac{1}{x-1}+\frac{-x-1}{x^2+9}\right)\ dx=\int\left(\frac{1}{x-1}-\frac{x+1}{x^2+9}\right)\ dx\\\\=\int\frac{1}{x-1}\ dx-\int\frac{x+1}{x^2+9}\ dx=\int\frac{1}{x-1}-\int\frac{x}{x^2+9}\ dx-\int\frac{1}{x^2+9}\ dx=(**)\\\\\#1\ \int\frac{1}{x-1}\ dx\Rightarrow\left|\begin{array}{ccc}x-1=t\\dx=dt\end{array}\right|\Rightarrow\int\frac{1}{t}\ dt=lnt+C_1=ln(x-1)+C_1$

$\#2\ \int\frac{x}{x^2+9}\ dx\Rightarrow \left|\begin{array}{ccc}x^2+9=u\\2x\ dx=du\\x\ dx=\frac{1}{2}\ du\end{array}\right|\Rightarrow\int\left(\frac{1}{2}\cdot\frac{1}{u}\right)\ du=\frac{1}{2}\int\frac{1}{u}\ du\\\\\\=\frac{1}{2}ln(u)+C_2=\frac{1}{2}ln(x^2+9)+C_2$

$\#3\ \int\frac{1}{x^2+9}\ dx=\int\frac{1}{x^2+3^2}\ dx=\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3\\\\therefore:\\\\\#1;\ \#2;\ \#3\Rightarrow(**)=ln(x-1)+C_1-\frac{1}{2}ln(x^2+9)+C_2-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3$

$\boxed{=ln(x-1)-\frac{1}{2}ln(x^2+9)-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C}$

4 0

3 years ago