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Angelina_Jolie [31]
3 years ago
12

Consider the quadratic function shown in the table below. x y 0 0 1 3 2 12 3 27 Which exponential function grows at a faster rat

e than the quadratic function for 0

Mathematics
2 answers:
olganol [36]3 years ago
8 0

Answer:

Graph D last graph

Step-by-step explanation:

ch4aika [34]3 years ago
3 0

Answer:

y=3x^2 is the quadratic that presents the table

x |   0           1           2           3

y |   0           3          12          27

The exponential that includes (1,4) takes off a bit faster than the others near x=0.

Step-by-step explanation:

If the table is:

x |   0           1           2           3

y |   0           3          12          27

It says it is a quadratic and we know the y-intercept is 0 since we have the point (0,0).

y=ax^2+bx

Now we need to use 2 more mores and we should wind up with a system to solve for a and b.

(1,3)                                                              (2,12)

3=a(1)^2+b(1)                                                12=a(2)^2+b(2)

3=a+b                                                           12=4a+2b

                                                                     6=2a+b  (divided both sides by 2)

So we are solving the system

3=a+b

6=2a+b

-------------- I'm going to choose elmination because it is already setup that way. I'm going to subtract the equations.  In doing so, the b's will cancel.

3-6=a-2a

-3=-a

a=3

so since 3=a+b and a=3 then b=0

So the quadratic is y=3x^2

And perhaps we could have looked for an easier way to solve this since we winded up with such a simple quadratic. I'm saying it might have been just easier looking for a pattern but something it won't be that easy.

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Explain how to write a quadratic equation given the following three points on the graph (5,31) (3,11) (0,11)
masha68 [24]

Given:

The graph of a quadratic function passes through the points (5,31) (3,11) (0,11).

To find:

The equation of the quadratic function.

Solution:

A quadratic function is defined as:

y=ax^2+bx+c            ...(i)

It is passes through the point (0,11). So, substitute x=0,y=11 in (i).

11=a(0)^2+b(0)+c

11=c

Putting c=11 in (i), we get

y=ax^2+bx+11               ...(ii)

The quadratic function passes through the point (5,31). So, substitute x=5,y=31 in (ii).

31=a(5)^2+b(5)+11

31-11=a(25)+5b

20=25a+5b

Divide both sides by 5.

4=5a+b                  ...(iii)

The quadratic function passes through the point (3,11). So, substitute x=3,y=11 in (ii).

11=a(3)^2+b(3)+11

11-11=a(9)+3b

0=9a+3b

Divide both sides by 3.

0=3a+b                 ...(iv)

Subtracting (iv) from (iii), we get

4-0=5a+b-3a-b

4=2a

\dfrac{4}{2}=a

2=a

Putting a=2 in (iv), we get

0=3(2)+b

0=6+b

-6=b

Putting a=2,b=-6 in (ii), we get

y=(2)x^2+(-6)x+11

y=2x^2-6x+11

Therefore, the required quadratic equation is y=2x^2-6x+11.

7 0
3 years ago
Math answer with work cause it need proof ASAP<br> for points
Sauron [17]

Answer:

A

Step-by-step explanation:

We are starting with 51

50x10=50

1x1=1

since there were 0.21 seconds, we first have to find the tenths place.

0.1x2=0.2

now that we found the tenths place, we need to find the hundredths place

0.01x1=1

FINAL ANSWER

(5 x 10) + (1 x 1) + (2 x 0.1) + (1 x 0.01)

5 0
4 years ago
Solve the equation. Then check your solution. – two-fifths + a = One-fifth a. Three-fifths c. StartFraction 3 Over 10 EndFractio
VLD [36.1K]

Answer:

a = Negative one-fifth

Step-by-step explanation:

The given equation is :

\dfrac{2}{5}+a=\dfrac{1}{5}

We need to find the value of a.

Subtract 2/5 from both sides.

\dfrac{2}{5}+a-\dfrac{2}{5}=\dfrac{1}{5}-\dfrac{2}{5}\\\\a=\frac{1}{5}-\frac{2}{5}\\\\a=\dfrac{-1}{5}

So, the correct answer is Negative one-fifth. Hence, the correct option is (b).

5 0
3 years ago
How much did each pack of yarn cost?? 22 points
Law Incorporation [45]

Answer:

$13

Step-by-step explanation:

89 = 11 +6x

78 = 6x

x = 13

6 0
3 years ago
Is this correct? help please
Drupady [299]
Yes! That graph matches the equation
7 0
3 years ago
Read 2 more answers
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