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klemol [59]
3 years ago
10

Solve for x 10x = 3 Give your answer as a fraction in it’s simplest form.

Mathematics
2 answers:
Illusion [34]3 years ago
4 0
Solve for X you have to divide the 3 by 10. You get 3/10 which is already in the simplest terms.
Blababa [14]3 years ago
4 0

Answer:

x = 3/10

Step-by-step explanation:

10x = 3

x = 3/10

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The vertices of a backyard are W(10,30),X(10,100),Y(110,100),Z(50,30). The coordinates are measured in feet. The line segment XZ
MariettaO [177]

Given vertices W(10,30), X(10,100), Y(110,100), Z(50,30) of a backyard, find distances WX, XY, YZ, ZW and XZ:

1. WX=\sqrt{(10-10)^2+(30-100)^2}=\sqrt{70^2}=70\ ft;

2. XY=\sqrt{(110-10)^2+(100-100)^2}=\sqrt{100^2}=100\ ft;

3. YZ=\sqrt{(110-50)^2+(100-30)^2}=\sqrt{60^2+70^2}=10\sqrt{85}\ ft;

4. ZW=\sqrt{(50-10)^2+(30-30)^2}=\sqrt{40^2}=40\ ft;

5. XZ=\sqrt{(10-50)^2+(100-30)^2}=\sqrt{40^2+70^2}=10\sqrt{65}\ ft.

Then:

1. the area

A_{WXZ}=\sqrt{\frac{70+40+10\sqrt{65}}{2}\\\cdot (\frac{70+40+10\sqrt{65}}{2}-70)\cdot (\frac{70+40+10\sqrt{65}}{2}-40)\cdot (\frac{70+40+10\sqrt{65}}{2}-10\sqrt{65})}=\\ \\=1400\ ft^2.

2. the area

A_{XYZ}=\sqrt{\frac{100+10\sqrt{85}+10\sqrt{65}}{2}\cdot (\frac{100+10\sqrt{85}+10\sqrt{65}}{2}-100)}\cdot\\ \\\cdot\sqrt{(\frac{100+10\sqrt{85}+10\sqrt{65}}{2}-10\sqrt{85})\cdot (\frac{100+10\sqrt{85}+10\sqrt{65}}{2}-10\sqrt{65})}=3500\ ft^2.

Then

\dfrac{A_{XYZ}}{A_{WXZ}}=\dfrac{3500}{1400}=2.5 times greater.

6 0
4 years ago
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