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Darya [45]
3 years ago
8

I need help with this question

Mathematics
1 answer:
Nastasia [14]3 years ago
8 0

Answer:

120 cm³

Step-by-step explanation:

V = (1/3)×base area×height

  = (1÷3)×(4×9)×10

  =120

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4. Identify the domain and range of the relation<br><br> {(2,4),(7,-9),(1,4),(3, -6), (8,0)}
Sergeeva-Olga [200]

Answer:

Domain: (2,7,1,3,8) Range: (4,-9,-6,0)

Step-by-step explanation:

4 0
2 years ago
Aaron needs 24 inches of copper wire for an experiment. The wire is sold by centimeter. How many centimeters of wire does Aaron
vovikov84 [41]

Answer:

60.96 cm

Step-by-step explanation:

you do proportions

1 in = 2.54

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4 0
3 years ago
Read 2 more answers
PLEASE PEOPLE, HELP ME!! Geometry
Oksanka [162]
I use the sin rule to find the area

A=(1/2)a*b*sin(∡ab)

1) A=(1/2)*(AB)*(BC)*sin(∡B)
sin(∡B)=[2*A]/[(AB)*(BC)]

we know that
A=5√3
BC=4
AB=5
then

sin(∡B)=[2*5√3]/[(5)*(4)]=10√3/20=√3/2
(∡B)=arc sin (√3/2)= 60°

 now i use the the Law of Cosines 

c2 = a2 + b2 − 2ab cos(C)

AC²=AB²+BC²-2AB*BC*cos (∡B)

AC²=5²+4²-2*(5)*(4)*cos (60)----------- > 25+16-40*(1/2)=21

AC=√21= 4.58 cms

the answer part 1) is 4.58 cms

2) we know that

a/sinA=b/sin B=c/sinC

and

∡K=α

∡M=β

ME=b

then

b/sin(α)=KE/sin(β)=KM/sin(180-(α+β))

KE=b*sin(β)/sin(α)

A=(1/2)*(ME)*(KE)*sin(180-(α+β))

sin(180-(α+β))=sin(α+β)

A=(1/2)*(b)*(b*sin(β)/sin(α))*sin(α+β)=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)

A=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)

KE/sin(β)=KM/sin(180-(α+β))

KM=(KE/sin(β))*sin(180-(α+β))--------- > KM=(KE/sin(β))*sin(α+β)

the answers part 2) are

side KE=b*sin(β)/sin(α)
side KM=(KE/sin(β))*sin(α+β)
Area A=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)

5 0
2 years ago
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Alisiya [41]

Answer:

The area of the parallelagram is 50

5 0
3 years ago
Cameron had 30 minutes to do a three problem quiz. He spent 8 3/4 minutes on question A and 5 1/2 minutes on question B. How muc
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He has 15 and 3/4 minutes left over
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2 years ago
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