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BabaBlast [244]
3 years ago
6

Any REAL help on #16 above??? Can someone check my answer?

Mathematics
1 answer:
babymother [125]3 years ago
3 0

Answer:

The answer to your question is:th first option is correct.

Step-by-step explanation:

Here we have and hyperbola with center (0, 1), and the hyperbola is horizontal because x² is positive.

Equation

y - k = ±\frac{b}{a} (x - h)

Process

Find a, b

                a² = 9    

               a = 3

                b² = 5

              b = √5

              h = 0    and k = 1

Substitution

              y - 1 = ±\frac{√5}{3} (x - 0)

Equation 1

            y = \frac{\sqrt{5} }{3} x + 1

Equation 2

           y = - \frac{\sqrt{5} }{3} x + 1  

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If square root of x has a decimal approximation of 3.4, then x is between which two integers? In your final answer, include all
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Answer:

You know that √x = 3.4, therefore x = 3.4² = 11.56

Hope I helped :)

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In which quadrant does the point (13, 18) lie
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Answer:

Quadrant 1

Step-by-step explanation:

how I understood this is first understanding how a graph works. Once you understand that you can count 13 horizontal on the "X" Axis, then go 18 up on the "Y" axis. you should get a point in the upper left area and that area is called Quadrant 1. Say if 13 was negative instead of going horizontal east you would go horizontal west. And the point should end up in Quadrant 2. Any more questions feel free to ask me below.

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1. There are 4 quarts in 1 gallon. How many quarts are in 4 gallons?
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vitfil [10]

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i think its 67 (im not sure)

<em />

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Rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the
Brut [27]

Answer:

The time that the rocket will hit the ground is 12.87 seconds.

Step-by-step explanation:

Given : The rocket is launched from a tower  y=-16x^2+199x+90. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation.

To find : The time that the rocket will hit the ground ?

Solution :

When the rocket hit the ground i.e. height became zero y=0,

Equation is y=-16x^2+199x+90

Substitute y=0,

-16x^2+199x+90=0

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x=\frac{-(199)\pm\sqrt{(199)^2-4(-16)(90)}}{2(-16)}

x=\frac{-199\pm\sqrt{45361}}{-32}

x=\frac{-199+\sqrt{45361}}{-32},\frac{-199-\sqrt{45361}}{-32}

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Reject negative value.

The time taken is 12.87 seconds.

5 0
3 years ago
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