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2 years ago

8

Help pls fast it is urgent

1 answer:

quester [9]2 years ago

5 0

Answer:

The volume formula is pi r^2h because it is similar to the formula to the area of a circle with an extra addition of the height variable, since a cylinder is just a circle with depth. The surface area formula is 2pirh +2pi r^2 because it measures the area of both circles as well as the depth. This formula is similar to the circumference formula of a circle.

Step-by-step explanation:

Mark me brainliest pls

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If f is a function, then f(3x)=3f(x)

Delvig [45]

This is false.
The f(3x) means you are multiplying the input 'x' by 3 before plugging into the equation.
3f(x) means you are going to multiply your answer (output) by 3 after you've plugged the input value in and gotten the output.

7 0

3 years ago

-k(-4k^2+k+2) Best one and simplest one gets the brainliest, I'm In a rush.

sasho [114]

Answer:

4k^3 - k^2 - 2k

Step-by-step explanation:

-k * -4k^2 = 4k^3

-k * k = -k^2

-k * 2 = -2k

4k^3 - k^2 - 2k

4 0

3 years ago

Read 2 more answers

Help!!<br>n/4+7=8<br>(/ means fraction)

Harman [31]

THE CORRECT ANSWER IS N=4

8 0

2 years ago

The half-life of a radioactive element is 130 days, but your sample will not be useful to you after 80% of the radioactive nu

gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt} =-\lambda N$

$\Rightarrow \frac{dN}{N} =-\lambda dt$

Integrating both sides

$\int \frac{dN}{N} =\int\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$\Rightarrow ln|N_0|=-\lambda .0+c$

$\Rightarrow c= ln|N_0|$

$\therefore ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$ .......(1)

$\frac{N}{N_0}=e^{-\lambda t}$ .........(2)

Logarithm:

$ln|\frac mn|= ln|m|-ln|n|$
$ln|ab|=ln|a|+ln|b|$

$ln|e^a|=a$

$ln|a|=b \Rightarrow a=e^b$
$ln|1|=0$

130 days is the half-life of the given radioactive element.

For half life,

$N=\frac12 N_0$ , $t=t_\frac12=130$ days.

we plug all values in equation (1)

$ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130$

$\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130$

$\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130$

$\rightarrow -ln|2|=-\lambda \times 130$

$\rightarrow \lambda= \frac{-ln|2|}{-130}$

$\rightarrow \lambda= \frac{ln|2|}{130}$

We need to find the time when the sample remains 80% of its original.

$N=\frac{80}{100}N_0$

$\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t$

$\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}$

$\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}$

$\Rightarrow t\approx 42$

We can use the sample about 42 days.

7 0

3 years ago

1. Rahul started his homework at 5.25 and ended his homework at 6:40. Find the elapsed time.

Darya [45]

Answer:

1. 75 minutes

2. 5:50

Step-by-step explanation:

you have to add and subtract

8 0

2 years ago