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3 years ago

PLEASE I NEED HELP ASAP {WILL REWARD BRAINLIST}

Mathematics

2 answers:

Elina [12.6K]3 years ago

5 0

The answer is g(x) = 2^x+3-2

ASHA 777 [7]3 years ago

4 0

Answer: D

Step-by-step explanation:

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The ninth graders are hosting the next school dance. They would like to make at least a $500 profit from selling tickets. The ni

kykrilka [37]

Answer:

They would earn 900 dollars at the door

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3 years ago

PLS HELP WILL GIVE BRAINLIEST IF CORRECT

ozzi

Answer:

Letting x represent the number of signatures

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520 + 6x = 1000

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3 years ago

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The difference of 22 and a number is -8 A.14 B.30 C.-30 D.-14

DIA [1.3K]

Answer:

B. 30

Step-by-step explanation:

the equation looks like 22 - x = -8 and 22 - 30 is negative 8

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3 years ago

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Adrianna sold her DVD that cost $20 with an 8% markup. How much did Adrianna make with the 8% markup?

djverab [1.8K]

Let’s find 1% of the markup first.
20 x 0.01 = $0.20
$0.20 x 8 = $1.60 markup.
$20 + $1.60 = $21.60 total.

Adrianna made $21.60 total, including markup.

Hope that helped! :)

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

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3 years ago