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4 years ago

PLEASE ANSWER ASAP FOR BRAINLIEST

Mathematics

1 answer:

saul85 [17]4 years ago

3 0

Answer:

Step-by-step explanation:

When you count the units it's 3, but because the scale factor is 4, you have to multiply it by that. 3 times 4 is 12.

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Percent decrease from 20 to 11

nevsk [136]

Answer:

45 percent

Step-by-step explanation:

20-11=9

9 divided by 20= 0.45

0.45 x 100= 45%

4 0

3 years ago

Sides of a square is 3^2/7 what is the area of the square

nydimaria [60]

Area = square of the measure of a side = 3^(2/7) * 3^(2/7)

= 3 ^(2/7 + 2/7) = 3^(4/7)

Option 2

6 0

3 years ago

10(y+2)-y=2(9y-8) i need to know how to solve it

vlabodo [156]

10(y+2)-y=2(9y-8)

Mutiply the first bracket by 10

Mutiply the second bracket by 2

(10)(y)(10)(2)= 10y+20

(2)(9y)(2)(-8)= 18y-16

10y+20-y= 18y-16

10y-y+20= 18y-16

9y+20= 18y-16

move 18y to the other side

sign changes from +18y to -18y

9y-18y+20= 18y-18y-16

-9y+20= -16

move +20 to the other side

sign changes from +20 to -20

-9y+20-20=-16-20

-9y= -36

divide by -9

-9y/-9= -36/-9

y= 4

Answer: y= 4

6 0

3 years ago

Which is the mode of the data given below 24686424664

Gnom [1K]

The mode is: 6 and 4

7 0

3 years ago

After the mill in a small town closed down in 1970, the population of that town started decreasing according to the law of expon

wlad13 [49]

Answer:

$P_o = \frac{143000}{e^{-20*0.01303024661}}=110193.69$

And we can round this to the nearest up integer and we got 110194.

Step-by-step explanation:

The natural growth and decay model is given by:

$\frac{dP}{dt}=kP$ (1)

Where P represent the population and t the time in years since 1970.

If we integrate both sides from equation (1) we got:

$\int \frac{dP}{P} =\int kdt$

$ln|P| =kt +c$

And if we apply exponentials on both sides we got:

$P= e^{kt} e^k$

And we can assume $e^k = P_o$

And we have this model:

$P(t) = P_o e^{kt}$

And for this case we want to find $P_o$

By 1990 we have t=20 years since 1970 and we have this equation:

$143000 = P_o e^{20k}$

And we can solve for $P_o$ like this:

$P_o = \frac{143000}{e^{20k}}$ (1)

By 2019 we have 49 years since 1970 the equation is given by:

$98000 = P_o e^{49k}$ (2)

And replacing $P_o$ from equation (1) we got:

$98000=\frac{143000}{e^{20k}} e^{49k} =143000 e^{29k}$

We can divide both sides by 143000 we got:

$\frac{98000}{143000} =0.685 = e^{29k}$

And if we apply ln on both sides we got:

$ln(0.685) = 29k$

And then k =-0.01303024661[/tex]

And replacing into equation (1) we got:

$P_o = \frac{143000}{e^{-20*0.01303024661}}=110193.69$

And we can round this to the nearest up integer and we got 110194.

7 0

3 years ago