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3 years ago

Exterior Angles and Isosceles Triangles

Mathematics

2 answers:

Naddika [18.5K]3 years ago

7 0

Exterior angles =180-75=105
Interior angles =180_69=111

Ad libitum [116K]3 years ago

3 0

Step-by-step explanation:

this is not an isoceles triangle.

anyway, we need to remember 2 things :

1. the sum of all angles in a triangle is always 180°.

2. the sum of all angles around a single point on a line is 360° (as this point wound be the center of an invisible circle). and as the line cuts this circle in half, the angles on one side of the line around this point is therefore 180°.

so, the internal angle Q is then

180 = 50 + 55 + Q

Q = 180 - 50 - 55 = 180 - 105 = 75°

the exterior angle x at Q is then

180 = Q + x = 75 + x

x = 180 - 75 = 105°

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marissa [1.9K]

Answer:

y + 1 = 3(x - 1)

Step-by-step explanation:

The given point is (1,-1) meaning that x1 = 1 and y1 = -1

The slope of this line is m = 3 because we go up 3 and over to the right 1 (eg: go from the point (1,-1) to (2,2) to see this in action)

Plug these three pieces of info into the point slope formula below

y - y1 = m(x - x1)

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

y + 1 = 3(x - 1)

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3 years ago

Read 2 more answers

(-7.5) + (-5.3) (-1.9)

Lisa [10]

Answer:

2.57

Step-by-step explanation:

You first have to multiply the -5.3 times the -1.9. Once you do that, you get 10.07. Then, you add the -7.5.

Here are the equation steps:

(-7.5) + (-5.3)x(-1.9)

(-7.5) + (10.07)

2.57

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3 years ago

Help pls don't answer if you don't know also I’ll give brainliest

il63 [147K]

Evaluate for x=4,y=4,z=4

|(2)(4)−4+(3)(4)|

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Answer=16

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3 years ago

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denis-greek [22]

Let $X$ be the random variable denoting the number of girls that are born. Then $X$ has a binomial distribution on $n=125$ babies and probability of getting a girl $p=0.5$ . The probability we want is

$\mathbb P(X\ge60)=\displaystyle\sum_{x=60}^{125}\binom{125}x0.5^{125-x}0.5^x\approx0.704=70.4\%$

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3 years ago

Read 2 more answers

Match the expressions with their equivalent simplified expressions.

Tasya [4]

Answer:

$\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}}\rightarrow\frac{2x}{3y}\\\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} \rightarrow\frac{3y}{2x}\\\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}}\rightarrow\frac{4x^2}{5y}\\\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}}\rightarrow\frac{3x^2}{2y}\\\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} \rightarrow\frac{2xy}{3}\\\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}}\rightarrow\frac{x}{2y}$

Step-by-step explanation:

$\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}} =\sqrt[4]{\frac{(2^4)(x^{6-2})(y^{4-8})}{(3^4)}} =\sqrt[4]{\frac{2^4x^4y^{-4}}{3^4}} =\frac{2xy^{-1}}{3}=\frac{2x}{3y}$

$\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} =\sqrt[4]{\frac{(3^4)(x^{2-6})(y^{10-6})}{(2^4)}} =\sqrt[4]{\frac{3^4x^{-4}y^{4}}{2^4}} =\frac{3x^{-1}y^1}{3}=\frac{3y}{2x}$

$\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}} =\sqrt[3]{\frac{(4^3)(x^{8-2})(y^{7-10})}{(5^3)}} =\sqrt[3]{\frac{4^3x^6y^{-3}}{5^3}} =\frac{4x^2y^{-1}}{5}=\frac{4x^2}{5y}$

$\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}} =\sqrt[5]{\frac{(3^5)(x^{17-7})(y^{16-21})}{(2^5)}} =\sqrt[5]{\frac{3^5x^{10}y^{-5}}{2^5}} =\frac{3x^2y^{-1}}{2}=\frac{3x^2}{2y}$

$\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} =\sqrt[5]{\frac{(2^5)(x^{12-7})(y^{15-10})}{(3^5)}} =\sqrt[5]{\frac{2^5x^{5}y^{5}}{3^5}} =\frac{2x^1y^{1}}{3}=\frac{2xy}{3}$

$\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}} =\sqrt[4]{\frac{(2^4)(x^{10-2})(y^{9-17})}{(4^4)}} =\sqrt[4]{\frac{2^4x^{8}y^{-8}}{4^4}} =\frac{2x^{1}y^{-1}}{4}=\frac{x}{2y}$

Thus,

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3 years ago