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Gwar [14]
2 years ago
14

The graph of a polynomial function has zeroes 0 (multiplicity 2) , 1 (multiplicity 2), and 5/2 (multiplicity 2). Write a functio

n in standard form that could represent this function
Mathematics
1 answer:
pogonyaev2 years ago
6 0

Answer:

x^6 - 7x^5 + 69/4x^4 + -35/2x^3 + 25/4x^2

Step-by-step explanation:

x^2(x-1)^2(x-5/2)^2 is the factored form of the polynomial. Now you simplify it to convert to standard form.

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Find the value of x and y is (3x_6) (12,y-1)​
Alika [10]

Answer:

x=4

y=7

Step-by-step explanation:

12=3x

x=4

6=y-1

y=6+1

y=7

8 0
3 years ago
Read 2 more answers
How to put 5/27 into a <br> percentage
leonid [27]

Answer:

18.5185%

Step-by-step explanation:

(7/27) * 100% = (0.185185) * 100% = 18.5185%

5 0
2 years ago
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Find the standard form of the equation of the parabola with a focus at (0, 8) and a directrix at y = -8.
Gala2k [10]
Vertex is directly in middle of directix and focus

distance from 8 to -8 is 16
16/2=8
so 8 below focus (since 8>-8) is the point (0,0
vertex is (0,0)
nice

it opens up because focus is above directix
also it goes up down so
4p(y-k)=(x-h)^2
(h,k) is veretx
we got that (h,k) is (0,0)
and p is distance from vertex to focus which is 8
so
4(8)(y-0)=(x-0)^2
32y=x^2
y=(1/32)x^2
5 0
3 years ago
Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$
Viktor [21]

When n=0, we have

5^{3^0} + 1 = 5^1 + 1 = 6

3^{0 + 1} = 3^1 = 3

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for n=k, that

3^{k + 1} \mid 5^{3^k} + 1

Now for n=k+1, we have

5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)

so we know the left side is at least divisible by 3^{k+1} by our assumption.

It remains to show that

3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1

which is easily done with Fermat's little theorem. It says

a^p \equiv a \pmod p

where p is prime and a is any integer. Then for any positive integer x,

5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3

Furthermore,

5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3

which goes all the way down to

5^{3^k} \equiv 5 \pmod 3

So, we find that

\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3

QED

5 0
1 year ago
0.5w+3x,when w=6,x=5
Nina [5.8K]

Answer:

3 + 15 =

18

Step-by-step explanation:

0.5 x 6 =3

3 x 5 = 15

5 0
2 years ago
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