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2 years ago

Is 8/12 greater than 10/15? and how do you know?

Mathematics

2 answers:

valentina_108 [34]2 years ago

8 0

Answer:

no, they are equal. we know this because when reducing both they are equal.

Step-by-step explanation:

first find the least common denominator

that's 3

reduce fractions until the denominator is both 3

we get 2/3 and 2/3. the proportions are equal.

VARVARA [1.3K]2 years ago

8 0

8/12 is .6667, and 10/15 is also .6667. They are the same decimals, which means they are the same value. To answer your question, No, 8/12 is not greater than 10/15, they are exactly the same

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If W(-10, 4), X(-3, -1), and Y(-5, 11) classify ΔWXY by its sides. Show all work to justify your answer.

solniwko [45]

Given:

The vertices of ΔWXY are W(-10, 4), X(-3, -1), and Y(-5, 11).

To find:

Which type of triangle is ΔWXY by its sides.

Solution:

Distance formula:

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

Using distance formula, we get

$WX=\sqrt{(-3-(-10))^2+(-1-4)^2}$

$WX=\sqrt{(-3+10)^2+(-5)^2}$

$WX=\sqrt{(7)^2+(-5)^2}$

$WX=\sqrt49+25}$

$WX=\sqrt{74}$

Similarly,

$XY=\sqrt{\left(-5-\left(-3\right)\right)^2+\left(11-\left(-1\right)\right)^2}=2\sqrt{37}$

$WY=\sqrt{\left(-5-\left(-10\right)\right)^2+\left(11-4\right)^2}=\sqrt{74}$

Now,

$WX=WY$

So, triangle is an isosceles triangles.

and,

$WX^2+WY^2=(\sqrt{74})^2+(\sqrt{74})^2$

$WX^2+WY^2=74+74$

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$WX^2+WY^2=(2\sqrt{37})^2$

$WX^2+WY^2=WY^2$

So, triangle is right angled triangle.

Therefore, the ΔWXY is an isosceles right angle triangle.

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