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gulaghasi [49]
3 years ago
7

Which numbers are a distance of 3 units from 12 on a number line? A number line ranging from negative 5 to 17. Select each corre

ct answer.

Mathematics
2 answers:
spin [16.1K]3 years ago
4 0

The distance between A and B is d = |B - A|

We have d = 3 and A = 12.

Substitute:

|B-12|=3\iffB-12=3\ \vee\ B-12=-3\qquad|\text{add 12 to both sides}\\\\B=15\ \vee\ B=9

<h3>Answer: 9 and 15.</h3>

aliya0001 [1]3 years ago
3 0

Answer:

(11) (-3) i took the test

Step-by-step explanation:

please mark me as brainlyist

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What is the domain of f?
Sedbober [7]

The domain of a function is the set of all possible inputs for the function. For example, the domain of f(x)=x² is all real numbers, and the domain of g(x)=1/x is all real numbers except for x=0.

5 0
2 years ago
Express in kg: 3470 grams<br>4 kg 8 g​
Airida [17]

Answer:

Step-by-step explanation:

This Makes no sense but this is 4kg and 8g converted to Kg :

4kg - 8g = 4.008 kg

8 0
3 years ago
a radioactive sample contains 3.00 g of pure 11c, which has a half-life of 20.4 min. (a) how many moles of 11c are present initi
sveta [45]

The number of mole of pure 11C (carbon-11) initially present is 0.2727 mole

<h3>Description of mole </h3>

The mole of a substance is related to it's mass and molar mass according to the following equation

Mole = mass / molar mass

With the above formula, we can obtain the number of mole of pure 11C. Detail below:

<h3>How to determine the mole of 11C</h3>

From the question given, the following data were obtained:

  • Molar mass of 11C = 11 g/mol
  • Mass of 11C = 3 g
  • Mole of 11C =?

Mole = mass / molar mass

Mole = 3 / 11

Mole = 0.2727 mole

Therefore, the number of mole of pure 11C initially present is 0.2727 mole

Learn more about mole:

brainly.com/question/13314627

#SPJ4

8 0
1 year ago
Graphing rational functions
Debora [2.8K]

Answer:

Opinion B

Step-by-step explanation:

hope that will help you

5 0
2 years ago
Find an equation of the plane through the point (−5,−1,2) with normal vector ????=⟨−1,−5,2⟩.
Ganezh [65]

Answer:

x + 5y - 2z + 14 = 0

Step-by-step explanation:

A plane that passes through the point (x_{0}, y_{0}, z_{0}) with a normal vector of (a,b,c) has the following equation, initially:

a(x - x_{0}) + b(y - y_{0}) + c(z - z_{0}) = 0

After we solve this, we have

ax + by + cz + d = 0

In this problem, we have that:

(x_{0}, y_{0}, z_{0}) = (-5,-1,2)

(a,b,c) = (−1,−5,2)

So

a(x - x_{0}) + b(y - y_{0}) + c(z - z_{0}) = 0

-1(x - (-5)) - 5(y - (-1)) + 2(z - 2) = 0

-(x + 5) - 5(y + 1) + 2(z - 2) = 0

-x - 5 - 5y - 5 + 2z - 4 = 0

-x - 5y + 2z - 14 = 0

Multplying everything by -1

x + 5y - 2z + 14 = 0

3 0
3 years ago
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