I NEED HELP ON STORY PROBLEM!!

I need the answer ASAP please

BartSMP [9]

Answer:

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░░░▌░▄▄▄▐▌▀▀▀░░ This is Bob

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▀█▌░░░▄░▀█▀░▀ ░░ Copy And Paste Him onto all of ur brainly answers

░░░░░░░▄▄▐▌▄▄░░░ So, He Can Take

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Step-by-step explanation:

4 0

3 years ago

Figure out the costs of buying the two cars listed below by filling in the blanks in the table

a_sh-v [17]

Make/Model:
Ford E-Series Wagon Van

MSRP: 
$28,760

Cost of options: 
$5,560

Sales tax: 
$2230.80 (this is based on 6.5% interest)

Total cost: 
$36550.80

10% down payment: 
$3655.08

Amount needed to borrow: 
$32895.72

Monthly payment: 
$665.62

Total interest paid: 
$4,665.51

Total payments:
60 payments of total $39,937.03

6 0

3 years ago

Members of a baseball team raise $967.50 to go to a tournament. They rented a bus for $450.00 and budgeted $28.75 per player for

WINSTONCH [101]

967.50 = 450 + 28.75p
967.50 - 450 = 28.75p
517.5 = 28.75p
18 = p

7 0

3 years ago

Read 2 more answers

The measure of an angle is 51 degrees

ki77a [65]

Supplementary angle = 180

180 - 51 = x

x = 129

C) 129 degrees is your answer

hope this helps

5 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