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2 years ago

(3, 4) was dilated to become point X' (12, 16). What was the scale factor?

Mathematics

1 answer:

tensa zangetsu [6.8K]2 years ago

6 0

Step one: correspond x values and y values together

it would be: 3, 12 and 4, 16

Step two: divide

12/3 = 4 and 16/4 = 4

both numbers = 4

scale factor is 4

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3 years ago

Which of the values shown are potential roots of f(x) = 3x3 – 13x2 – 3x + 45? Select all that apply.

galina1969 [7]

Answer:

All potential roots are 3,3 and $-\frac{5}{3}$ .

Step-by-step explanation:

Potential roots of the polynomial is all possible roots of f(x).

$f(x)=3x^3-13x^2-3x+45$

Using rational root theorem test. We will find all the possible or potential roots of the polynomial.

p=All the positive/negative factors of 45

q=All the positive/negative factors of 3

$p=\pm 1,\pm 3,\pm 5\pm \pm 9,\pm 15\pm 45$

$q=\pm 1,\pm 3$

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$\frac{p}{q}=\pm 1,\pm 3,\pm 5\pm \pm 9,\pm 15\pm 45,\pm \frac{1}{3},\pm \frac{5}{3}$

Now we check each rational root and see which are possible roots for given function.

$f(1)= 3\times 1^3-13\times 1^2-3\times 1+45\Rightarrow 32\neq 0$

$f(-1)= 3\times (-1)^3-13\times (-1)^2-3\times (-1)+45\Rightarrow \neq 32$

$f(-3)= 3\times (-3)^3-13\times (-3)^2-3\times (-3)+45\Rightarrow \neq -144$

$f(3)= 3\times (3)^3-13\times (3)^2-3\times (3)+45\Rightarrow =0\\\\ \therefore x=3\text{ Potential roots of function}$

Similarly, we will check for all value of p/q and we get

$f(-5/3)=0$

Thus, All potential roots are 3,3 and $-\frac{5}{3}$ .

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</em>
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