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Crank
4 years ago
8

Ron brought cupcakes to school to

Mathematics
2 answers:
mixer [17]4 years ago
7 0

Answer:

3/8

Step-by-step explanation:

8/8-5/8= 3/8

Schach [20]4 years ago
4 0

Answer:The fraction of the cupcakes that Ron have left to take home is 3/8

Step-by-step explanation:

Let x represent the total amount of cupcakes that Ron brought to school. By the end of the day, 5/8 of the cupcakes were gone. This means that the total mount of cupcakes that Ron had already shared is 5/8 × x = 5x/8.

The amount of cupcakes left would be x - 5x/8 = (8x - 5x)/8 = 3x/8

The fraction of the cupcakes that Ron have left to take home would be the amount that he has left divided by the initial amount. It becomes

(3x/8)/x = 3x/8 × 1/x = 3/8

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Answer:

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Step-by-step explanation:

Check answer:

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3 years ago
Solve the compound inequality and write the solution in interval notation: 34x−3≤3 or 25(x+10)≥0.
kirza4 [7]

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Step-by-step explanation:

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3 years ago
Find the difference quotient and simplify your answer. f(x) = 4x − x2, f(4 + h) − f(4) h , h ≠ 0
Anon25 [30]

The difference quotient and simplification will be    = [4 -h-2x]

The given equation is as follows:   f(x)= 4x - x²

For finding the quotient and further simplification we must follow the following steps:

[f(x + h) - f(x)] / h = [4(x + h) - (x + h)² - 4x + x²]/ h

<h3>What is simplification of algebraic operations?</h3>

Getting the functions in their lowest terms is known as simplification.

Brackets will get open and solved further;

[f(x + h) - f(x)] / h = [4(x + h) - (x + h)² - 4x + x²]/ h

[f(x + h) - f(x)] / h = [4h - h² - 2x]/ h  

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7 0
1 year ago
Use the exponential decay​ model, Upper A equals Upper A 0 e Superscript kt​, to solve the following. The​ half-life of a certai
Akimi4 [234]

Answer:

It will take 7 years ( approx )

Step-by-step explanation:

Given equation that shows the amount of the substance after t years,

A=A_0 e^{kt}

Where,

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Then if t = 19, amount of the substance = \frac{A_0}{2},

i.e.

\frac{A_0}{2}=A_0 e^{19k}

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Taking ln both sides,

\ln(0.5) = \ln(e^{19k})

\ln(0.5) = 19k

\implies k = \frac{\ln(0.5)}{19}\approx -0.03648

Now, if the substance to decay to 78​% of its original​ amount,

Then A=78\% \text{ of }A_0 =\frac{78A_0}{100}=0.78 A_0

0.78 A_0=A_0 e^{-0.03648t}

0.78 = e^{-0.03648t}

Again taking ln both sides,

\ln(0.78) = -0.03648t

-0.24846=-0.03648t

\implies t = \frac{0.24846}{0.03648}=6.81085\approx 7

Hence, approximately the substance would be 78% of its initial value after 7 years.

5 0
3 years ago
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