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Gnesinka [82]
3 years ago
15

Which formula should be used to find the area of a trapezoid​

Mathematics
2 answers:
frutty [35]3 years ago
5 0

Answer:

h(b1+b2)/2

Step-by-step explanation:

since it's not a full rectangle the height needs to be divided by 2.

yawa3891 [41]3 years ago
4 0
You have the correct chosen
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Help, plz............
Brilliant_brown [7]

Answer:

I think Question (a) is 11748 m and Question (b) is 32364

Step-by-step explanation:

Because I Just divided the number of 7031 m square by 79 in Question (a) and I Just multiplied the number of 348 m by 93 but I am not sure It's correct or not

I hope It's helpful

6 0
3 years ago
Please help with these questions... I've been trying to figure them out for like twenty minutes and at this point I just don't h
vlada-n [284]

Answer:

1. y=2x^2-8X-1

2. y=-6x^2+36x-63

3. y=-3x^2+18x-34

4. y=9x^2+54x+76

5. y=-x^2+8x-22

6. y=6x^2-48x+90

7. y=8x^2+16x+5

8. y=2x^2-8x+1

Have a great day :)

3 0
2 years ago
Could I get some help?
galina1969 [7]
The answer is x=5
Hope this helps
4 0
3 years ago
Find the area of this regular polygon.<br> Round to the nearest tenth.<br> 8.65 mm<br> [? ]mm2
nadezda [96]

Answer:

Actually it's not polygon. it's a nonagon. With r=8.65mm″, the law of cosines gives us side a:

a=√{b²+c²−2bc×cos40°}

a=√{149.645−149.645cos40°}

Area Nonagon = (9/4)a²cos40°

=9/4[149.645−149.645cos40°]cot20°

=336.70125[1−cos(40°)]cot(20°)

Applying an identity for the cos(40°) does not get us very far…

= 336.70125[1−(cos2(20°)−1)]cot(20°)

= 336.70125[2−cos2(20°)]cot(20°)

= 336.70125[2−(1−sin2(20°))]cot(20°)

= 336.70125[1+sin2(20°)]cos(20°)sin(20°)

= 336.70125[cot(20°)+sin(20°)cos(20°)]mm²

3 0
3 years ago
Choose one of the squares facts from problem 1?
egoroff_w [7]
I think you forgot to put in the image
8 0
3 years ago
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