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sergiy2304 [10]
3 years ago
9

Solve the equation. 5^3x-1 = 5^5 x=?

Mathematics
2 answers:
Jet001 [13]3 years ago
8 0

x=25

Step-by-step explanation:

evaluate the power 125x=5^5

125=3125

divide both sides by 125

x=25

leonid [27]3 years ago
4 0

Answer:

x= 3126/125 or x=25.008 or as a mixed number form: x=25 and 1/125

Step-by-step explanation:

isolate the variable by dividing each side by factors that don't contain the variable.

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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
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 now i use the the Law of Cosines 

c2 = a2 + b2 − 2ab cos(C)

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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)

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One way of doing this would be to look at the πr² as if it were a single term. Then we could divide both sides by πr², which leaves h = V/πr².

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