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Blizzard [7]
4 years ago
6

What is the lowest value of the range of the function shown on the graph?

Mathematics
1 answer:
IrinaK [193]4 years ago
4 0

Answer:

2

Step-by-step explanation:

2

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10+6*(75*50/100)-4/8
sergiy2304 [10]

First, do what is in the parentheses. 75 times 50 is 3,750. Divide it by 100 to get 37.5. 10 + 6(37.5) - 1/2. Do the multiplication to get 10 + 225 - 1/2. Now you get 235 - 1/2. Solve to get 234 1/2.



7 0
3 years ago
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Please someone help me with this asap! Giving brainliest !!!!!!!!!
Lina20 [59]

Answer:

I thank its b not sure sorry if its wrong

3 0
3 years ago
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How do I solve this problem??
Luda [366]

Answer:

3/10 m^2

Step-by-step explanation:

To calculate area of a triangle we multiply height with base and then divide that by 2

2/5 × 3/2 ÷ 2 = 3/10

3 0
3 years ago
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
3 years ago
What is 696969 time 1
AVprozaik [17]

Anything times 1 = the same number so 696969x1=696969

3 0
3 years ago
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