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Naddika [18.5K]

3 years ago

6

HELPPPP!!!! geometry plzzz..... What is the distance between (3,-4) and (8,8) ?

1 answer:

kirza4 [7]3 years ago

4 0

Answer:

13

Step-by-step explanation:

To find the distance, d, between two points on the Cartesian plane use the distance formula.

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

The points are (3, -4) and (8, 8), so you can use:

x1 = 3

y1 = -4

x2 = 8

y2 = 8

Now plug in all values above in the formula and evaluate it.

$d = \sqrt{(8 - 3)^2 + [8 - (-4)]^2}$

$d = \sqrt{(5)^2 + (8 + 4)^2}$

$d = \sqrt{25 + (12)^2}$

$d = \sqrt{25 + 144}$

$d = \sqrt{169}$

$d = \sqrt{13}$

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If -1 is one of the zeroes of py 3x^3-5x^2-11x-3 find other zeroes

katrin2010 [14]

Given:

The function is:

$y=3x^3-5x^2-11x-3$

It is given that -1 is a zero of the given function.

To find:

The other zeroes of the given function.

Solution:

If c is a zero of a polynomial P(x), then (x-c) is a factor of the polynomial.

It is given that -1 is a zero of the given function. So, $(x+1)$ is a factor of the given function.

We have,

$y=3x^3-5x^2-11x-3$

Split the middle terms in such a way so that we get (x+1) as a factor.

$y=3x^3+3x^2-8x^2-8x-3x-3$

$y=3x^2(x+1)-8x(x+1)-3(x+1)$

$y=(x+1)(3x^2-8x-3)$

Again splitting the middle term, we get

$y=(x+1)(3x^2-9x+x-3)$

$y=(x+1)(3x(x-3)+1(x-3))$

$y=(x+1)(3x+1)(x-3)$

For zeroes, $y=0$ .

$(x+1)(3x+1)(x-3)=0$

$(x+1)=0$ and $3x+1=0$ and $x-3=0$

$x=-1$ and $x=-\dfrac{1}{3}$ and $x=3$

Therefore, the other two zeroes of the given function are $-\dfrac{1}{3}$ and $3$ .

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What is the average rate of change of the amount of the element over the 10-minute experiment? *

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Answer:

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Step-by-step explanation:

Given: a graph

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