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3 years ago

Find the following for the given f(x) and g(x) F(x)=2x+3;g(x)=3x+1

Mathematics

1 answer:

zheka24 [161]3 years ago

8 0

Answer:

this is questions in class 10ICSE mathematics chapter linear equations

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neptune is 4,498,252,900 kilometers from the sun. Which distance is 200,000,000 kilometers closer to the sun?

Virty [35]

The average distance of Neptune from the Sun is 2,795,084,800 miles or 4,498,252,900 kilometers. Because its orbit is elliptical, its distance from the Sun changes depending on where it is in its orbit. The closest Neptune gets to the Sun is 2,771,087,000 miles or 4,459,630,000 kilometers. The farthest it gets from the Sun is 2,819,080,000 miles or 4,536,870,000 kilometers.

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3 years ago

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How do you estimate 289 and 7

padilas [110]

You first round 289 and get 290 then multiply it by 7

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2 years ago

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MArishka [77]

Answer:

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Step-by-step explanation:

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3 years ago

Elizabeth can jog 1 1/2 miles in 1/2 hour. How many miles can she jog in 1 hour

Nadya [2.5K]

Answer:

3 miles

Step-by-step explanation:

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3 years ago

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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

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3 years ago