Order the quadratic functions from widest to narrowest graph. y=4x^2 y=2x^2 y= 0.15x^2

Answers

Answer 1
Answer: ANSWER

y=0.15x^2, y=2x^2,y=4x^2

EXPLANATION

The given quadratic functions are;

y = 4 {x}^(2)
y = 2 {x}^(2)
and

y = 0.15 {x}^(2)

A fractional coefficient indicates a horizontal stretch.

The smaller the coefficient, the wider the stretch for
|a|  < 1
and

a \ne0

Also
|a|  > 1
indicates a horizontal compression.

The higher the value the narrower the graph.


Therefore from widest to narrowest, we have,

y=0.15x^2, y=2x^2,y=4x^2

See graph.

Related Questions

Order -3.98, 3 8/9, -3 11/12, and 3.9 repeated from least to greatest
Solve the proportional equation below: 4/7=28/(x+18)
jerome bought 2 postcards and received $1.35 in change in quarters and dimes. If he got coins back, how many of each coin did he get?
Find the area of the figure shown.
5+3=4d please help me

The population P of a bacteria culture is modeled by P = 4100e^kt where t is the time inhours. If the population of the culture was 5800 after 40 hours, how long does it take for
the population to double? Round to the nearest tenth of an hour.

Show work please
A LOT OF POINTS

Answers

Enter the given values into the equation and solve.

5800 = 4100e^(k*40)

Divide both sides by 4100 and simplify:

58 / 41 = e^(k*40)

Remove e by taking the logarithm of both sides:

ln(58/41) = k *40

Divide both sides by 40:

k = ln(58/41)/40

k = 0.00867

Now for the population to double set up the equation:

2*4100 = 4100e^kt

The 4100 cancels out on both sides:

2 = e^kt

Take the logarithm of both sides:

ln(2) = k*t

Divide both sides by k

t = ln(2) /k

replace k with the value from above:

t = ln(2) / 0.00867

t = 79.95

Rounded to the nearest tenth = 80.0 hours to double.

Answer:

It would take around 122 hours to double the population.

Step-by-step explanation:

To answer the question, we first need to find the constant k, using the given information and the expression.

P=4100e^(kt) \n5800=4100e^(k(40)) \n(5800)/(4100)=e^(40k)\ne^(40k)=1.41\nlne^(40k)=ln1.41\n40k=0.34\nk=(0.34)/(40)\approx 0.0085

Now that we have the constant. We can find the time it would take to double the population which would be 11600:

P=4100e^(kt)\n11600=4100e^(0.0085t)\n(11600)/(4100)= e^(0.0085t)\ne^(0.0085t)=2.83\nlne^(0.0085t)=ln2.83\n0.0085t=1.04\nt=(1.04)/(0.0085)\approx 122.35

Therefore, it would take around 122 hours to double the population.

Evaluate the expression ​

Answers

Answer:

47

Step-by-step explanation:

In the first step, inside the brackets I took the LCM and then in second step I subtracted 3 from 11 which gave result of8÷8=1 then 1 is multiplied to 2 which becomes 49-2 and 47 is the result

3 hundreds, 17 tens, 3 ones, 1 tenths, and 2 hundredths. what number am I

Answers

473.12 because you have to add 300 plus 170 (17 tens) plus 3 plus .1 plus .02

How many solutions does the equation y=3y+5 have

Answers

it has one solution y =--2.5

Need helpp fastt plss will give brainliest to who is correct plss helpp

Answers

I think I might have my math wrong here but I think 5

Which of the following answer choices are correct ?

Answers

Okay, here we go. 

So, given the formula, we can just plug in the numbers on the number lines. So for example:

On the first graph, it starts at 4 and goes to 5. Let's plug those in:
-6(4)+42<24
-24+42<24
18<24
18 is less than 24, so this is true so far.

moving on:
-6(5)+42<24
-30+42<24
12<24
true, again. The first graph seems correct. Let's check our answers by using the third graph where the numbers 3,2, and 1 are checked.

-6(3)+42<24
-18+42<24
24<24
24 is equal to 24, not less than, so the third graph is wrong. 

Knowing this, we can use the process of elimination. We know 4 and 5 fit the equation and 3, 2, and 1 don't. This means that the graph with 4 and 5 marked is the correct one aka the first graph

Hopefully this is correct and I helped at all :)