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ollegr [7]
3 years ago
13

Does anyone wanna talk

Mathematics
2 answers:
gayaneshka [121]3 years ago
7 0
Sure what we talking about
Cause right now I need 2 brainlys to rank up o_O
kvasek [131]3 years ago
6 0

Answer:

dadating din yan❤ hehehe

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Whats y-2=6x+3 in y=mx+b form
Lynna [10]
Adding 2 to the three gives you y=6x+5
4 0
4 years ago
Read 2 more answers
Only answer the part where it says her bonus will be______
luda_lava [24]
Her bonus would be four dollars. I hope this helped!
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3 years ago
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243.875 to nearest tenth, hundredth, ten, and hundred
IgorC [24]
243.875 to the nearest tenth = 243.9
243.875 to the nearest hundredth = 243.88
243.875 to the nearest ten = 240
243.875 to the nearest hundred = 200
This number is going to be a little bit trickier. The basic law of rounding is that is the number is less than 5, it is going to stay the same. If the number is greater than or equal to 5 then it rounds up to the next number.
8 0
3 years ago
This is algebra 2 it’s growth & decay. I need help
Elodia [21]
<h3>Part 1:</h3>

So firstly, remember that the <em>a variable is the initial value.</em> Since we start off with 6.08 billion, <u>6.08 billion is the a variable.</u>

Next, since this is a <em>growth by 1.26%</em>, add 1 and 0.0126 (1.26% in decimal form) together to get 1.0126. <u>1.0126 is the b variable.</u>

Putting it together, <u>the function is A(t)=6.08(1.0126)^t</u>

<h3>Part 2:</h3>

Since 2020 is 20 years from 2000, plug in 20 into the t variable of the prior function to solve:

A(20)=6.08(1.0126)^{20}\\ A(20)=7.81

In context, <u>in 2020 the population will be 7.81 billion.</u>

6 0
4 years ago
I am confused on numbers 25 and 29, the instructions are at the top.
kvasek [131]
#25 is fairly simple.  Plug in -4 and 3 into the equation, and the extraneous root will be the one that does not work.
\sqrt{12-(-4)} =  \sqrt{16} =  \frac{+}{}4
Extraneous root in this case is positive four since +4≠-4
<span>\sqrt{12-3} = \sqrt{9} = \frac{+}{}3
</span>In this case it's negative 3, since -3≠3

#29 can be turned into a quadratic equation.
x= \sqrt{2x+3}
Square both sides to get
x^{2}=2x+3
Then bring the 2x+3 to the other side, setting the quadratic equal to zero.
x^{2}-2x-3=0
Factor to find that it's equivalent to
(x-3)(x+1)=0
Therefore x is equal to positive 3 and negative 1.  Plug both back into the original equation.  Whichever does not work is the extraneous root, and the answer is the one that does.
<span>x= \sqrt{2x+3}
</span><span>3= \sqrt{2(3)+3}
</span><span>3= \sqrt{9}
</span>Extraneous root would be negative 3.

<span>-1= \sqrt{2(-1)+3}
</span><span>-1= \sqrt{1}
</span>Extraneous root would be positive 1.

Your answers are positive 3 and negative 1.
Extraneous roots are negative 3 and positive 1.
3 0
3 years ago
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