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2 years ago

9

VJKL has vertices J (6,2), K (1,3), and

1 answer:

klio [65]2 years ago

8 0

Answer:

r3234r32r

Step-by-step explanation:

dfefedwfdsf4r36557tgrfdgdgfgreagf

dfefedwfdsf4r36557tgrfdgdgfgreagf

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HELP HELP HELP HELP HELP HELP<br> PLEASE

Brilliant_brown [7]

Answer: 1 $\frac{5}{27}$

Step-by-step explanation:

$\frac{8}{9} x 1\frac{1}{3} = 1\frac{5}{27}$

3 0

2 years ago

Read 2 more answers

In the standard (x,y) coordinate plane a right triangle has vertices at (-3,4)(3,4)(3,-4) what is the length in coordinates unit

MAXImum [283]

Given:

The vertices of a right triangle are:

$(-3,4),(3,4),(3,-4)$

To find:

The length of the hypotenuse of the given right triangle.

Solution:

Let the vertices of the right triangle are $A(-3,4),B(3,4),C(3,-4)$ .

The distance formula is:

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

Using distance formula, we get

$AB=\sqrt{(3-(-3))^2+(4-4)^2}$

$AB=\sqrt{(3+3)^2+(0)^2}$

$AB=\sqrt{(6)^2+0}$

$AB=\sqrt{36}$

$AB=6$

Similarly,

$BC=\sqrt{(3-3)^2+(-4-4)^2}$

$BC=\sqrt{(0)^2+(-8)^2}$

$BC=\sqrt{64}$

$BC=8$

And,

$AC=\sqrt{(3-(-3))^2+(-4-4)^2}$

$AC=\sqrt{(6)^2+(-8)^2}$

$AC=\sqrt{36+64}$

$AC=\sqrt{100}$

$AC=10$

Now, taking sum of squares of two smaller sides, we get

$AB^2+BC^2=6^2+8^2$

$AB^2+BC^2=36+64$

$AB^2+BC^2=100$

$AB^2+BC^2=AC^2$

By the definition of the Pythagoras theorem, AC is the hypotenuse of the given triangle.

Therefore, the length of the hypotenuse is 10 units.

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3 years ago

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vivado [14]

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3 years ago

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brilliants [131]

First, ask somebody else like your teacher, also use your brain

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2 years ago

You randomly select one card from a 52-card deck. Find the probability of selecting a black three or a red jack ) The probabilit

matrenka [14]

1/13

2 black 3s and 2 red Jacks, 4 cards out of 52.

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2 years ago