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3 years ago

When 12 is subtracted from 3 times the number the result is 24 what's the number

1 answer:

3 0

$x-\ number\\\\
\frac{12}{3x}=24\ \ \ \ | multiply\ by\ 3x\\\\
12=24*3x\\\\
12=72x\ \ \ \ | divide\ by\ 12\\\\
x=6\\\\Number\ is\ 6.$

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F(x) = x? – x3 Select the value of f(-2) 4 -12 12 A B C D

Talja [164]

Step-by-step explanation:

f(x) = x² - x³

f(-2) = (-2)² - (-2)³

= 4 - (-8)

= 4 + 8 = 12

3 0

3 years ago

CAN SOMEONE PLEASE HELP ME

Paraphin [41]

Answer:

Line CDE is a straight line,

so 85°+70°=55°

180°-55°=25°

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3 years ago

Write the improper fraction 51/8 as a mixed number

garri49 [273]

Answer:

6 3/8

Step-by-step explanation:

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3 years ago

Read 2 more answers

Pictures is bad but find length bc

aleksley [76]

Answer:

BC=5

Step-by-step explanation:

4/10=x/x+3

4(x+3)=10x

4x+12=10x

subtract 4x from each side

6x=12

divide both sides by 6

6x/6=12/6

x=2

BC=x+3

substitute for x

BC=2+3

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5 0

3 years ago

Suppose water is added to a tank at 10 gal / min , but leakes out at the rate of 0.2 gal / min for each gallon in the tank . Wha

Nana76 [90]

Answer:

50 gallon

Step-by-step explanation:

We are given that

$\frac{dV_{in}}{dt}=10 gal/min$

Let V(in gallon) be the present volume of tank.

$\frac{dV_{out}}{dt}=0.2 V gal/min$

We have to find the smallest capacity the tank can have if the process is to continue indefinitely.

According to question

$\frac{dV}{dt}$ =Incoming volume-outgoing volume

$\frac{dV}{dt}=10-0.2 V=-(\frac{1}{5}V-10)=-(\frac{V-50}{5})$

$\frac{5dV}{V-50}=-dt$

Integrating on both sides

$\int \frac{5dV}{V-50}=-\int dt$

$5ln(V-50)=-t+ln C$

By using the formula

$\int \frac{dx}{x}=ln x+C$

$ln(V-50)=\frac{1}{5}(-t+ln C)$

$ln(V-50)=-\frac{1}{5}t+\frac{1}{5} ln C$

$ln(V-50)=-\frac{1}{5} t+lnC^{\frac{1}{5}}$

$ln(V-50)=-\frac{1}{5}t+ln C'$

Where $ln C'=ln C^{\frac{1}{5}}$

$ln(V-50)-ln C'=-\frac{1}{5} t$

$ln\frac{V-50}{C'}=-\frac{1}{5}$

Using the formula

$ln m-ln n=ln\frac{m}{n}$

$\frac{V-50}{C'}=e^{-\frac{1}{5}t}$

$V-50=C'e^{-\frac{1}{5}t}$

Substitute $t=\infty$

$V-50=0$

$V=50$

Hence, the smallest capacity the tank can have if the process is to continue indefinitely=50 gallon

7 0

3 years ago