What is the equation that passes through the point(-4,-4) and has a slope of 1/6

Save Work out the size of angle AED.<br> b) Work out x

Mamont248 [21]

Answer:

x = 10°

Step-by-step explanation:

a). Since, opposite angles of a cyclic quadrilateral are supplementary angles"

Therefore, in cyclic quadrilateral ABDE,

m∠ABD + m∠AED = 180°

110° + m∠AED = 180°

m∠AED = 180° - 110°

= 70°

b). AD = ED [Given]

m∠EAD = m∠AED [Since, opposite angles of equal sides are equal in measure]

m∠EAD = m∠AED = 70°

By triangle sum theorem in ΔABD,

m∠BAD + m∠ABD + m∠ADB = 180°

m∠BAD + 110° + 40° = 180°

m∠BAD = 180 - 150

= 30°

m∠AEB = m∠AED + m∠DAB [By angles addition postulate]

m∠AEB = 70° + 30°

= 100°

By triangle sum theorem in the large triangle,

x° + m∠AEB + m∠EAB = 180°

x° + 100° + 70° = 180°

x = 180 - 170

x = 10°

8 0

3 years ago

Find the volume of the pyramid.

xz_007 [3.2K]

I think it is 432.sorry if it is wrong though. My bad calculation error.

3 0

3 years ago

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,

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

3 0

3 years ago

What is the equation of the line shown in this graph?

Grace [21]

Answer:

x=-2

Step-by-step explanation:

It is a straight line when x=-2 :)

8 0

3 years ago

Two competing sports equipment suppliers sell footballs at different prices. Supplier A charges $85 in shipping, and charges $2.

S_A_V [24]

Supplier A: $188.6
Supplier B: $221.6

I hope I helped but I honestly have no idea if I did. Anyway sorry if I didn't. Good luck.

7 0

3 years ago