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Nimfa-mama [501]
3 years ago
14

What is the range of this graph?

Mathematics
2 answers:
torisob [31]3 years ago
4 0
Can’t see the graph, try reposting it
Lelechka [254]3 years ago
3 0
Did you forgot to put your picture in the question
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Are any solutions possible for this problem..
dimulka [17.4K]

The equation: -2(x+3)=-2x-6, is simplified with distribution of parentheses: -2x-6=-2x-6.

Now add 2x to both sides and get: -6=-6.

The numbers are equal therefore the solution to the equation x can be any number x\in\mathbb{R}.

8 0
3 years ago
miley's music has a sale on music cds. all music cds are discounted 15%. mariana's receipt indicates that she saved 3$ on her cd
velikii [3]


The answer is 45.

You have to multiply 15% by $3.

When you multiply them you get 45.


5 0
4 years ago
Which of the following values is not in the domain of y = tanx?
SSSSS [86.1K]

Answer:

I think its the first option

Step-by-step explanation:

I'm not sure

7 0
3 years ago
Read 2 more answers
An old bone contains 80% of its original carbon-14. Use the half-life model to find the age of the bone. Find an equation equiva
Ira Lisetskai [31]

Answer:

An old bone contains 80% of its original carbon-14 in 1844.6479 years

Step-by-step explanation:

We know that

half life time of C-14 is 5730 years

so, h=5730

now, we can use formula

P(t)=P_0(\frac{1}{2})^{\frac{t}{h} }

we can plug back h

and we get

P(t)=P_0(\frac{1}{2})^{\frac{t}{5730} }

An old bone contains 80% of its original carbon-14

so,

P(t)=0.80Po

we can plug it and then we solve for t

0.80P_0=P_0(\frac{1}{2})^{\frac{t}{5730} }

0.80=(\frac{1}{2})^{\frac{t}{5730} }

\ln \left(0.8\right)=\ln \left(\left(\frac{1}{2}\right)^{\frac{t}{5730}}\right)

t=-\frac{5730\ln \left(0.8\right)}{\ln \left(2\right)}

t=1844.6479

So,

An old bone contains 80% of its original carbon-14 in 1844.6479 years


4 0
3 years ago
Read 2 more answers
What are the answers to these questions?
allochka39001 [22]

Answerplz someone answer i need help on this

Step-by-step explanation:

7 0
3 years ago
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