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2 years ago

What is the LCD for the given equation?

Mathematics

1 answer:

velikii [3]2 years ago

5 0

Answer:

(x-3)^2

Step-by-step explanation:

The LCD is (x-3)^2 This is becuse x-3 is on both equations.

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For f(x)=1/x^2-3, find a) f(3) b) f(2+h) Show/explain work

valentina_108 [34]

$f(x)=\dfrac{1}{x^2-3}\\\\a)\ f(3)\to\text{put x=3 to the equation of the function f}\\\\f(3)=\dfrac{1}{3^2-3}=\dfrac{1}{9-3}=\dfrac{1}{6}\\\\Answer:\ \boxed{f(3)=\dfrac{1}{6}}\\\\b)\ f(2+h)\to\text{put x=2+h to the equation of a function f}\\\\f(2+h)=\dfrac{1}{(2+h)^2-3}\ \ \ |\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\=\dfrac{1}{2^2+2(2)(h)+h^2-3}=\dfrac{1}{4+4h+h^2-3}=\dfrac{1}{h^2+4h+1}\\\\Answer:\ \boxed{f(2+h)=\dfrac{1}{h^2+4h+1}}$

6 0

3 years ago

Can someone help please Stickers are sold in packages of 15 sheets. If there are 20 students in the class, how many packages of

MrRa [10]

Answer:

4 packages

Step-by-step explanation:

If there are 20 and stickers are sold in 15 sheets, then you got to work backwards. take the answer and multiply it by the amount of sheets. For example: 4 times 15 is 60, 60 divided by 20 is 3. No leftover numbers.

8 0

2 years ago

Read 2 more answers

Which of the following is a sequence that moves on a basis of addition and subtraction?

irinina [24]

It is the first answer

5 0

3 years ago

What is the sum of 1 2/5 and 5 3/4

Art [367]

Answer:

$7\frac{3}{20}$

Step-by-step explanation:

Hey there!

Well to add this we need to pu it in improper form.

7/5 + 23/4

Now we need to find the LCM.

5 - 5, 10, 15, 20, 25, 30

4 - 4, 8, 12, 16, 20, 24, 28

So the LCD is 20.

Now we need to change the 5 and 4 to 20.

5*4 = 20

7*4 = 28

28/20

4*5=20

23*5=115

115/20

Now we can add 28 and 115,

= 143/20

Simplified

7 3/20

Hope this helps :)

4 0

3 years ago

Read 2 more answers

How much of a radioactive kind of thorium will be left after 14,680 years if you start with

babymother [125]

Answer:

8978 grams

Step-by-step explanation:

The equation to find the half-life is:

$N(t)= N_{0}e^{-kt}$

N(t) = amount after the time t

$N_{0}$ = initial amount of substance

t = time

It is known that after a half-life there will be twice less of a substance than what it intially was. So, we can get a simplified equation that looks like this, in terms of half-lives.

$N(t)= N_{0}e^{-\frac{ln(\frac{1}{2}) }{t_{h} } t}$ or more simply $N(t)= N_{0}(\frac{1}{2})^{\frac{1}{t_{h} } }$