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2 years ago

Given segment AE = 3x+5, BE = 4, DE = 5 and CE = 2x+6, what is the value of segment AE

Mathematics

1 answer:

dimulka [17.4K]2 years ago

6 0

Answer:

the answer is 3 I think but am not sure of the workings

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Please help me with this

Cerrena [4.2K]

Answer:

Yes;Each side of triangle PQR is the same length as the corresponding side of triangle STU

Step-by-step explanation:

You can observe the sides of both triangles to see if this property holds

Lets check the length of AB

A(0,3) and B(0,-1)

$AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\\\\\\\AB=\sqrt{(0-0)^2+(-1-3)^2} \\\\\\\\AB=\sqrt{0^2+-4^2} \\\\\\AB=\sqrt{16} =4units$

Now check the length of the corresponding side DE

D(1,2) and E(1,-2)

$DE=\sqrt{(1-1)^2+(-2-2)^2} \\\\\\DE=\sqrt{0^2+-4^2} \\\\\\DE=\sqrt{16} =4units$

The side AB has the same length as side DE.This is also true for the remaining corresponding sides.

6 0

3 years ago

Can someone please help me with geometry?????

crimeas [40]

Ok what is the 'q' ???

4 0

2 years ago

The perpendicular bisector of the line segment connecting the points (-3,8) and (-5,4) has an equation of the form y = mx + b. F

omeli [17]

Answer:

Step-by-step explanation:

find the slope

$\frac{4-8}{-5-(-3)} =\frac{-4}{-2} \\\\slope=2\\y=mx+b\\y=2x+b\\$

take a coordinate to fill in

$(-5,4)\\y=-5\\x=4\\-5=2(4)+b\\-5=8+b-8 -8\\-13=b\\$

this means that the equation is y=2x-13

and if you add m and b

you get :-11

<u><em>I HOPE THIS HELPS</em></u>

8 0

2 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,

3 0

3 years ago

Help me pleaseeeeeee

iogann1982 [59]

You start by finding two points on the line. In this case, (-4,1) and (-2,2) will do.

To get from (-4,1) to (-2,2), you need to go “up 1, right 2” which gives you a slope of m = 1/2

Next you need the b-value, which comes from the y-intercept of (0,3). The b-value is 3.

Putting the slope and b-value into y=mx+b, you have y = 1/2 x + 3.

4 0

2 years ago