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GuDViN [60]
3 years ago
13

A triangle has side lengths of 6,8, and 9 what type of triangle is it

Mathematics
1 answer:
Strike441 [17]3 years ago
5 0
9^2 < 6^2 + 8^2 

so its acute angled.
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If the endpoints of the diameter of a circle are (−8, −6) and (−4, −14), what is the standard form equation of the circle?
kondaur [170]

Equation of the circle is (x+6)^{2}+(y+10)^{2}=20.

Solution:

The endpoints of the diameter of a circle are (–8, –6) and (–4, –14).

Center of the circle = Mid point of the diameter

Mid point formula:

$P(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)

Here, x_1=-8, y_1=-6, x_2=-4, y_2=-14

$P(x, y) =\left(\frac{-8-4}{2}, \frac{-6-14}{2}\right)

$P(x, y) =\left(\frac{-12}{2}, \frac{-20}{2}\right)

$P(x, y) =(-6, -10)

Center of the circle = (–6, –10)

Radius is the distance between center and any endpoint of the diameter.

To calculate the radius using distance formula.

r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}

Here, x_1=-6, y_1=-10, x_2=-8, y_2=-6

r=\sqrt{\left(-8-(-6)\right)^{2}+\left(-6-(-10)}\right)^{2}}

r=\sqrt{(-8+6)^{2}+(-6+10)^{2}}

r=\sqrt{(-2)^{2}+(4)^{2}}

r=\sqrt{20} units

The standard form of the equation of a circle is

(x-a)^{2}+(y-b)^{2}=r^{2}, where (a, b) are center and r is the radius.

Here, center = (–6, –10) and r=\sqrt{20}

(x-(-6))^{2}+(y-(-10))^{2}={(\sqrt{20})} ^{2}

(x+6)^{2}+(y+10)^{2}=20

Equation of the circle is (x+6)^{2}+(y+10)^{2}=20.

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3 years ago
Write .139 as a percent.
Rama09 [41]

Answer:

13.9%

Step-by-step explanation:

0.139 x 100=13.9, can you give me brainiest if its correct

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Here's the picture With the work

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Fiandnd product 3/4 x 1 ,1/2
VMariaS [17]
3/4 times 1 1/2 = 1 1/8

Hope this helps!
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A chef cooked 7 kilograms of food for dinner. He ate 11/12 of the amount, how much did he eat?
Bogdan [553]

11/12 of 7 is simply their product.


\bf 7\cdot \cfrac{11}{12}\implies \cfrac{77}{12}\implies 6\frac{5}{12}

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