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3 years ago

What is 34 tens + 20 tens =

1 answer:

7 0

34 tens + 20 tens = ?

1 ten = 10 ones or 10.

34 + 20 = 54
54 tens
34 tens + 20 tens = 54 tens

54 × 10 = 540

34 tens + 20 tens = ?
34 tens + 20 tens = 54 tens
34 tens + 20 tens = 540

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Equation -1-7v+3= how do i slove this i dont know

TEA [102]

Negative one minus negative seven times volume plus three equals BLANK. If v is a variable, the question is wrong.

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3 years ago

PLEASE ANSWER ASAP!!!

Vesna [10]

Answer:

C=x (x+3)

Step-by-step explanation:

x cannot divide x+3 definitely so the denominators must be multiplied to get the least common denominator.

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3 years ago

What 4 coins equal 90¢

N76 [4]

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half dollar, quarter, dime and nickel

Step-by-step explanation:

When you sum it up it will give you 90 cents.

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4 years ago

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<img src="https://tex.z-dn.net/?f=5%20%2B%20%20%5Cfrac%7Bx%20-%202%7D%7B3%7D%20%20%3D%20%20%5Cfrac%7Bx%20%2B%203%7D%7B8%7D%20" i

sammy [17]

First multiply each term by 24(the common denominator found by 8×3) to remove the fractions and make things easier.
$24(5) + 24( \frac{x - 2}{3} ) = 24( \frac{x + 3}{8} )$
This will give you,
$120 + 8(x - 2) = 3(x + 3)$
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7 0

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