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kupik [55]
3 years ago
9

I have a LCM and GCF homework can someone help i dont understand??

Mathematics
1 answer:
Sindrei [870]3 years ago
6 0
Alright,

So LCM stands for "least common multiple"
While GCF is "greatest common factor"

Let's look at your number 1 (3&6)

3,6,9,12
6,12,18,24

Both times table has 6 in them making 6 the LCM

To find the GCF we need to know what can go into 3 equally. Since 3 is prime the greatest factor for 3 is....3. 6 goes into 3 two times so 3 is the GCF.

LCM:6
GCF:3

Lets do number 10 (40&4)

For LCM you see whats the smallest number that is in both times tables

4,8,12,16,20,24,28.......40
40,80, 160....

40 is the LCM because 40 is what's the smallest number between the two

GCF?

4 times 1 equals 4. Nothing bigger than 4 can make 4 (if your multiplying). That makes 4 automatically the GCF. :)
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Can someone help me please
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Answer:

(3/2)π units²

Step-by-step explanation:

Hello, Katherine,

Your central angle, Ф, is 60°.

To find the area of this sector, we recognize that its central angle, 60°, is (1/6) of the area of the entire circle:

A of sector = (1/6)(π)(3 units)²  = 9π units²

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The exponential decay function A = A0(1/2)^t/P can be used to determine the amount A, of a radioactive substance present at time
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The half-life of the substance is about 288 days.

Step-by-step explanation:

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\displaystyle A=A_0\left(\frac{1}{2}\right)^{t/P}

Can determine the amount <em>A</em> of a radioactive substance present at time <em>t. A₀ </em>represents the initial amount and <em>P</em> is the half-life of the substance.

We are given that a substance loses 70% of its radioactivity in 500 days, and we want to determine the period of the half-life.

In other words, we want to determine <em>P. </em>

Since the substance has lost 70% of its radioactivity, it will have only 30% of its original amount. This occured in 500 days. Therefore, <em>A</em> = 0.3<em>A₀</em> when <em>t</em> = 500 (days). Substitute:

\displaystyle 0.3A_0=A_0\left(\frac{1}{2}\right)^{500/P}

Divide both sides by <em>A₀:</em>

\displaystyle 0.3=\left(\frac{1}{2}\right)^{500/P}

We can take the natural log of both sides:

\displaystyle \ln(0.3)=\ln\left(\left(\frac{1}{2}\right)^{500/P}\right)

Using logarithmic properties:

\displaystyle \ln(0.3)=\frac{500}{P}\left(\ln\left(\frac{1}{2}\right)\right)

So:

\displaystyle \frac{500}{P}=\frac{\ln(0.3)}{\ln(0.5)}

Take the reciprocal of both sides:

\displaystyle \frac{P}{500}=\displaystyle \frac{\ln(0.5)}{\ln(0.3)}

Use a calculator:

\displaystyle P=\frac{500\ln(0.5)}{\ln(0.3)}\approx287.86

The half-life of the substance is about 288 days.

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2 years ago
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