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dybincka [34]
3 years ago
13

Can someone explain how I do these problems

Mathematics
1 answer:
Serga [27]3 years ago
3 0
The answer is d. the angle that is 140* and the angle it is right next to must equal 180. therefore you subtract 180-140 to get the measure of that interior angle. after doing that, you add the measure of the two interior angles to get the measure of
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Find the area of this shape. Round your answer to the nearest tenth
MrMuchimi

Answer:

The area of this figure is 31,29.

Step-by-step explanation:

We have to find also the area of the rectangle, then we will find the area of the circle and divide by 2 because here is a half circle.

area of the rectangle = b x h

area of the rectangle = 6 x 6

area of the rectangle = 36

diameter = 6

r = d : 2

r = 6 : 2

r = 3 ft

area of the circle = π x r²

area of the circle = 3,14 x 3 ft

area of the circle = 9,42 ft

area of half of circle = 9,42 : 2 = 4,71 ft

area of this figure = area of rectangle - area of half of the circle

area of this figure = 36 - 4,71

area of this figure = 31,29

So the area of this figure is 31,29.

I hope this is helpful. :)

3 0
3 years ago
( (−7.2) 2 − 6.4 ) × (1.8 + (−0.8))
Kipish [7]

Answer:

I dont know what you want me to do. But if you wanted it simpilfied its

−20.8.

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
What is the equation of the graphed line written in standard form
weqwewe [10]
Ax+by=c


First you'll find the equation of the graphed line in slope-intercept form, y = mx+b, and then from there we'll convert it into standard form, ax+by = c.
5 0
3 years ago
What is the area of this<br>irregular shape?<br>.6 ft<br>square feet<br>3 ft<br>8 ft<br>16 ft<br>​
Andrej [43]

Answer:

Step-by-step explanation:

area=16×8+1/2×3×6

=128+9

=137 sq. ft.

5 0
3 years ago
How do I do 8b(ii) ? Please help me thank you!
Mila [183]
Step One
======
Find the length of FO (see below)

All of the triangles are equilateral triangles. Label the center as O
FO = FE = sqrt(5) + sqrt(2)

Step Two
======
Drop a perpendicular bisector from O to the midpoint of FE. Label the midpoint as J. Find OJ

Sure the Pythagorean Theorem. Remember that OJ is a perpendicular bisector.

FO^2 = FJ^2 + OJ^2
FO = sqrt(5) + sqrt(2)
FJ = 1/2 [(sqrt(5) + sqrt(2)]                                           \
OJ = ??

[Sqrt(5) + sqrt(2)]^2 = [1/2(sqrt(5) + sqrt(2) ] ^2 + OJ^2
5 + 2 + 2*sqrt(10) = [1/4 (5 + 2 + 2*sqrt(10) + OJ^2
7 + 2sqrt(10) = 1/4 (7 + 2sqrt(10)) + OJ^2      Multiply through by 4
28 + 8* sqrt(10) = 7 + 2sqrt(10) + 4 OJ^2    Subtract 7 + 2sqrt From both sides
21 + 6 sqrt(10) = 4OJ^2   Divide both sides by 4
21/4 + 6/4* sqrt(10) = OJ^2
21/4 + 3/2 * sqrt(10) = OJ^2 Take the square root of both sides.
sqrt OJ^2 = sqrt(21/4 + 3/2 sqrt(10) )
OJ = sqrt(21/4 + 3/2 sqrt(10) )

Step three
find h
h = 2 * OJ
h = 2* sqrt(21/4 + 3/2 sqrt(10) ) <<<<<< answer.


7 0
3 years ago
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