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3 years ago

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Brennan is planting seeds. He has 1/5 of a pack of seeds and 5 flower pots. What fraction of a pack of seeds will be planted in

each pot if he wants to use the same amount of seeds in each pot?

1 answer:

Maru [420]3 years ago

8 0

(1/5)/(5)=(1/5)(1/5)=1/25 of a pack of seeds will be planted in each pot

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(4÷y+2) +(2÷y-3) =(6÷y+6)

Evgesh-ka [11]

<span>Simplifying 4(y + -3) = 6(y + 2) Reorder the terms: 4(-3 + y) = 6(y + 2) (-3 * 4 + y * 4) = 6(y + 2) (-12 + 4y) = 6(y + 2) Reorder the terms: -12 + 4y = 6(2 + y) -12 + 4y = (2 * 6 + y * 6) -12 + 4y = (12 + 6y) Solving -12 + 4y = 12 + 6y Solving for variable 'y'. Move all terms containing y to the left, all other terms to the right. Add '-6y' to each side of the equation. -12 + 4y + -6y = 12 + 6y + -6y Combine like terms: 4y + -6y = -2y -12 + -2y = 12 + 6y + -6y Combine like terms: 6y + -6y = 0 -12 + -2y = 12 + 0 -12 + -2y = 12 Add '12' to each side of the equation. -12 + 12 + -2y = 12 + 12 Combine like terms: -12 + 12 = 0 0 + -2y = 12 + 12 -2y = 12 + 12 Combine like terms: 12 + 12 = 24 -2y = 24 Divide each side by '-2'. y = -12 Simplifying y = -12</span>

3 0

3 years ago

Tell whether the angles are complementary or supplementary. Then find the value of x

luda_lava [24]

Answer:

Complementary, x=78

Step-by-step explanation:

90= (x-8)+20

complementary angles add up to 90 degrees.

3 0

3 years ago

Read 2 more answers

If anyone could just look at my recent question pls its super easy its just simplyfing

vichka [17]

Answer: K

Step-by-step explanation:

Ill do it

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3 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

4 years ago

Write and solve an equation for each problem situation. Define the variable

Simora [160]

Answer:

His weekly allowance is $5.50.

Step-by-step explanation:

12.03 - 6.53 = $5.50

I'm not 100% sure this is accurate. Good Luck!

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3 years ago