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yanalaym [24]
3 years ago
8

Which equation shows y=34x−52y=34x−52 in standard form?

Mathematics
1 answer:
ladessa [460]3 years ago
5 0
It would be A, most definitely. 
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How many feet? Thank you for your time.
kramer

Answer:

1 3/8 Feet

Step-by-step explanation:

Take 2 1/2 and subtract 1 1/8 which gives u 1 3/8

if u don't understand fractions u can always turn ur fractions into decimals that's what I did :)

6 0
3 years ago
I need help on 5 and 6
Blizzard [7]

5. 4 rows of circles with 4 columns of circles


6. A Hexagon with 4 circles

8 0
3 years ago
What is the perimeter of the figure? <br><br> A. 144 cm <br> B. 76 cm <br> C. 74 cm <br> D. 45 cm
tatiyna
76 cm would be your answer, so B.
8 0
3 years ago
Read 2 more answers
What is the <img src="https://tex.z-dn.net/?f=%5Csqrt%7B10%7D" id="TexFormula1" title="\sqrt{10}" alt="\sqrt{10}" align="absmidd
morpeh [17]
6.32455532 is what my calculator says.
4 0
3 years ago
Find the area of each regular polygon. Round your answer to the nearest tenth if necessary.
tatuchka [14]

*I am assuming that the hexagons in all questions are regular and the triangle in (24) is equilateral*

(21)

Area of a Regular Hexagon: \frac{3\sqrt{3}}{2}(side)^{2} = \frac{3\sqrt{3}}{2}*(\frac{20\sqrt{3} }{3} )^{2} =200\sqrt{3} square units

(22)

Similar to (21)

Area = 216\sqrt{3} square units

(23)

For this case, we will have to consider the relation between the side and inradius of the hexagon. Since, a hexagon is basically a combination of six equilateral triangles, the inradius of the hexagon is basically the altitude of one of the six equilateral triangles. The relation between altitude of an equilateral triangle and its side is given by:

altitude=\frac{\sqrt{3}}{2}*side

side = \frac{36}{\sqrt{3}}

Hence, area of the hexagon will be: 648\sqrt{3} square units

(24)

Given is the inradius of an equilateral triangle.

Inradius = \frac{\sqrt{3}}{6}*side

Substituting the value of inradius and calculating the length of the side of the equilateral triangle:

Side = 16 units

Area of equilateral triangle = \frac{\sqrt{3}}{4}*(side)^{2} = \frac{\sqrt{3}}{4}*256 = 64\sqrt{3} square units

4 0
3 years ago
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