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lubasha [3.4K]
4 years ago
7

*Please Help* (There are more than 2 answers!)

Mathematics
2 answers:
GalinKa [24]4 years ago
4 0
Look like both B and D are correct answers. because 23*400=9200 and 26*400=10400
hammer [34]4 years ago
3 0

Answer:

The correct options are B and D.

Step-by-step explanation:

If a line passing through two points then the equation of line is

y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)

Choose any two points from the given table.

Let the two points are (1,400) and (3,1200).

The equation of line is

y-400=\frac{1200-400}{3-1}(x-1)

y-400=400(x-1)

y-400=400x-400

y=400x

Steven's monthly balances is defined by the equation

y=400x

At x=22,

y=400(22)=8800

At x=23,

y=400(23)=9200

At x=24,

y=400(24)=9600

At x=26,

y=400(26)=10400

At x=27,

y=400(27)=10800

The points (23, 9,200) and (26, 10,400) lie on the line. Therefore the correct options are B and D.

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Answer:

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Step-by-step explanation:

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4 years ago
The function h(t) = –16t2 + 96t + 6 represents an object projected into the air from a cannon. The maximum height reached by the
jolli1 [7]
If you've started pre-calculus, then you know that the derivative of  h(t)
is zero where h(t)  is maximum.

The derivative is            h'(t) = -32 t  +  96 .

At the maximum ...        h'(t) = 0

                                       32 t = 96 sec

                                           t  =  3 sec . 
___________________________________________

If you haven't had any calculus yet, then you don't know how to
take a derivative, and you don't know what it's good for anyway.

In that case, the question GIVES you the maximum height.
Just write it in place of  h(t), then solve the quadratic equation
and find out what  't'  must be at that height.

                                       150 ft = -16 t²  +  96  t  +  6 

Subtract 150ft from each side:    -16t²  +  96t  -  144  =  0 .

Before you attack that, you can divide each side by  -16,
making it a lot easier to handle:

                                                         t²  -  6t  +  9  =  0

I'm sure you can run with that equation now and solve it.    
The solution is the time after launch when the object reaches 150 ft.
It's 3 seconds.  
(Funny how the two widely different methods lead to the same answer.)

The answer is from AL2006

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Step-by-step explanation:

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A.is the answer Hope this helps and sorry it's late
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