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3 years ago

What’s the value of n? 3^n=1/9

Mathematics

1 answer:

Pepsi [2]3 years ago

4 0

Answer:

n= -2

Step-by-step explanation:

3^n = 1/9

3^n = 1/(3^2)

3^n = 3^(-2)

n = -2

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How much orange juice at 45 a liter should be mixed with grapefruit juice worth 35 a liter to make 200 liters of a mixture worth

miss Akunina [59]

The amount of liters is 200,if the orange juice is x than grape juice. 200-x
45x+35(200-x)=200*40
45x+7000-35x=8000
10x=1000
x=100

5 0

3 years ago

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write a point-slope equation for the line that passes through the point (6,8) and is parallel to the line given by y=-5x+4

VashaNatasha [74]

Answer:

y -8 = -5(x -6)

Step-by-step explanation:

The point-slope form of the equation for a line is generally written ...

y -k = m(x -h)

for slope m and point (h, k).

The slope of your parallel line is the same as the slope of the reference line, -5. So your equation is ...

y -8 = -5(x -6)

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3 years ago

3x-8<23 and -4x+26>6

Papessa [141]

Answer:

3x-8<23 and -4x+26>6

x<31/3 x<5

Step-by-step explanation:

3x-8<23

Add 8 to both sides

3x<31

Then divide 3 to both sides

x<31/3

and -4x+26>6

Minus 26 to both sides

-4x>-20

You divide -4 to both sides

Keep in mind that everytime you divide a negative the inequality will change so > will be <

x<5

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3 years ago

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$

Therefore the half-life is between 600 and 700 years

5 0

3 years ago

IF ANYONE HELP ME OUT WITH THESE PROBLEMS ILL GIVE EXTRA POINTS AND BRAINLEST

Dennis_Churaev [7]

Answer:

it's obtuse. reason being cause to be acute it has to be under 90 and to be right it has to be at 90

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3 years ago

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