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Naddika [18.5K]
3 years ago
6

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

Mathematics
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)

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Again splitting the middle term, we get

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

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

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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

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Given: a graph

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