Which polynomial function has a leading coefficient of 1, roots –2 and 7 with multiplicity 1, and root 5 with multiplicity 2? f(x) = 2(x + 7)(x + 5)(x – 2) f(x) = 2(x – 7)(x – 5)(x + 2) f(x) = (x + 7)(x + 5)(x + 5)(x – 2) f(x) = (x – 7)(x – 5)(x – 5)(x + 2)

Option 4

Step-by-step explanation:

We can check the options one by one.

for f(x)=2(x+7)(x+5)(x-2)

We can see that

Putting the factors one by one equal to zero will give us:

The above roots are not according to the given requirements so it is not the correct answer.

For f(x)=2(x-7)(x-5)(x+2)

The roots will be:

The roots are not according to the requirements of the question as the multiplicity of 5 is not two

For f(x) = (x + 7)(x + 5)(x + 5)(x – 2)

The roots will be:

Again the roots are not according to given requirements

For f(x) = (x – 7)(x – 5)(x – 5)(x + 2)

For finding the roots, put the function equal to zero

We can see that the roots are -2, 7 and 5 and the root 5 has multiplicity of two.

For leading co-efficient the factors will be multiplied:

We can see that the leading co-efficient is 1.

So option 4 is the correct answer ..

answer: b)

step-by-step explanation:

its a dot plot because since the score is not high numbers.

- r3ktforgood ☕

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl