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MAVERICK [17]
2 years ago
14

Compare the functions below:

Mathematics
2 answers:
rjkz [21]2 years ago
6 0
f(x)\to y_{max}=4\\\\g(x)=4\cos(2x-\pi)-2\to y_{max}=2\\\\h(x)=-(x-5)^2+3\to y_{max}=3

Answer: a. f(x)
krek1111 [17]2 years ago
5 0

Answer:

Option: a is the correct answer.

a.  f(x)

Step-by-step explanation:

  • We are given a set of values for the function f(x)  as:

x     y =f(x)

0   −5

1     0

2     3

3     4

4     3

5     0

6      −5

Clearly from the set of values we could observe that:

The maximum value of the function f(x) is: 4

  • Now we are given function g(x) as:

g(x)=4 \cos(2x-\pi)-2

We know that maximum value of g(x) is attained when the cosine function attains the maximum value.

Also the maximum value of cosine function is: 1

Hence, the maximum value of g(x) is : 4-2=2

  • Now we are given a quadratic function h(x) as:

h(x)=-(x-5)^2+3

As we know that the function:

(x-5)^2\geq 0\\\\This\ implies\ that:\\\\-(x-5)^2\leq 0\\\\-(x-5)^2+3\leq 3

Hence, the maximum value of  function h(x) is: 3

             Hence, the function that has the largest maximum is:

                        f(x)  ( which is 4)

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What is the parent function of<br> f(x)=2/5(-x-5)^5 +2
mamaluj [8]

Answer:

Parent function: x^{5}

Step-by-step explanation:

Parent function is the simplest form of the type of function given

for equation: \frac{2}{5} (-x -5)^{5} +2

the simplest form is: x^{5}

4 0
3 years ago
Please help me with the below question.
VMariaS [17]

By letting

y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}

we get derivatives

y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}

y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0

Examine the lowest degree term \left(x^{r-1}\right), which gives rise to the indicial equation,

5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients c_k is

(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}

so that with r = 4/5, the coefficients are governed by

c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}

c) Starting with c_0=1, we find

c_1 = -\dfrac{c_0}5 = -\dfrac15

c_2 = -\dfrac{c_1}{10} = \dfrac1{50}

so that the first three terms of the solution are

\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}

4 0
2 years ago
Which equation shows the value of -2/5-(-9/15)
enot [183]

Answer:

D.

Step-by-step explanation:

-2/5 - (-9/15) = -2/5 + 9/15.

both of those equations are equivalent.

hope this helps.

7 0
2 years ago
Read 2 more answers
Maria and her friends ate 5/8 of the 16 oranges that were at Maria's house. which is a reasonable answer for 5/8 x 16? please he
neonofarm [45]

Answer:

10

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
-7 1 11 10 29<br> 6 3 6 3 6<br> common difference?
Hoochie [10]

9514 1404 393

Answer:

  the listed sequences have no common difference(s)

Step-by-step explanation:

The differences in the first sequence are ...

  8, 10, -1, 19

These values are not equal, so there is no common difference.

__

The differences in the second sequence are ...

  -3, 3, -3, 3

These values are not equal, so there is no common difference.

_____

The difference is the difference between a term value and the one before.

  1 -(-7) = 8

  3 -6 = -3

7 0
3 years ago
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