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3 years ago

Help me rn pleasee !!

Mathematics

2 answers:

ankoles [38]3 years ago

7 0

Answer:

There are two solutions for x;

x = 0

x = $-\frac{3}{2}$

Step-by-step explanation:

kap26 [50]3 years ago

4 0

Answer:

x = 0 or -3/2

Step-by-step explanation:

( 2 x + 3 ) ^2 − 6 x − 9 = 0 ---> distribute

4x^2 + 6x + 6x + 9 - 6x - 9 = 0 --> add the 9's together

4x^2 + 12x - 6x = 0 --> combine like terms

4x^2 + 6x = 0 --> common factor

2(2x^2 + 3x) = 0 --> divide by 2 on both sides

2x^2 + 3x = 0

then you should use the quadratic formula (can't add here or the explanation would take too long)

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A circular garden has a circumference of 138 yards. Lars is digging a straight line along a diameter of the garden at a rate of

PilotLPTM [1.2K]

Answer:

4 hours

Step-by-step explanation:

Circumference formula:

C = πd

Use the given value and find the diameter:

138 = 3.14d
r = 138/3.14
r = 43.94 ≈ 44 yards

Time required to dig across the garden:

44/11 = 4 hours

6 0

2 years ago

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nalin [4]

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2 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:

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