a cable 48 meters long is cut into two pieces so that one piece is 18 neters longer than the other. Find the lenght of each piec
1 answer:
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Answer:
Step-by-step explanation:
This is a Combination (as in permutation vs combination) question the symbol (n r) refers to "n choose r". This is sometimes written as nCr
i.e the question is asking you to find how many combinations each will yield when you chose r items from n item without repetition and order does not matter.
I will only do the first question for you and you can just follow the same steps to solve the rest of the questions.
Recall that
Consider question a) we are given (5 1) or ₅C₁
we can see that n = 5 and r = 1
If we substitute this into the formula:
₅C₁ = (5!) / [ (1!)(5- 1)!]
= (5!) / [ (5- 1)!]
= (5!) / (4!)
= (5·4·3·2·1) / (4·3·2·1)
= 5
hence ₅C₁ = 5
The complete question in the attached figure
we know that
(see the attached figure n 2 to understand the problem)
[the surface area of one prism]=2*[x*x]+2*[x*y]+2*[x*y]----> 2x²+4xy
[the surface area of the sculpture]=2*[5*x*y]+2*[3*x*x]+2*[3*x*y]--> 6x²+16xy
now
<span>JD says the surface area of the sculpture is 4 times the surface area of one prism
</span>[the surface area of the sculpture]=4*(2x²+4xy)---> 8x²+16xy
we compare the value that JD says with the real value
(8x²+16xy) > (6x²+16xy)
the value that JD says is <span>greater in comparison with the real value
</span>This is because <span>JD should also subtract the areas of eight hidden surfaces.
the answer is
</span>JD should also subtract the areas of eight hidden surfaces<span>
</span>
we know that
(see the attached figure n 2 to understand the problem)
[the surface area of one prism]=2*[x*x]+2*[x*y]+2*[x*y]----> 2x²+4xy
[the surface area of the sculpture]=2*[5*x*y]+2*[3*x*x]+2*[3*x*y]--> 6x²+16xy
now
<span>JD says the surface area of the sculpture is 4 times the surface area of one prism
</span>[the surface area of the sculpture]=4*(2x²+4xy)---> 8x²+16xy
we compare the value that JD says with the real value
(8x²+16xy) > (6x²+16xy)
the value that JD says is <span>greater in comparison with the real value
</span>This is because <span>JD should also subtract the areas of eight hidden surfaces.
the answer is
</span>JD should also subtract the areas of eight hidden surfaces<span>
</span>
Let
A (x1,y1)-------> (1,5)
B (x2,y2)------> (5,10)
C (x3,y3)------> (9,15)
D (x4,y4)------> (13,20)
find the slope
AB
m=(10-5)/(5-1)-----> m=5/4
find the slope
CB
m=(15-10)/(9-5)-----> m=5/4
find the slope
DC
m=(20-15)/(13-9)-----> m=5/4
so
<span>the relationship shown is linear
</span>
<span>to model the data with an equation find the equation of a line
</span>with m=5/4 and point A (1,5)
y-y1=m*(x-x1)-----> y-5=(5/4)*(x-1)----> y=1.25x-1.25+5-----> y=1.25x+3.75
the equation to model data is
y=1.25x+3.75
see the attached figure
A (x1,y1)-------> (1,5)
B (x2,y2)------> (5,10)
C (x3,y3)------> (9,15)
D (x4,y4)------> (13,20)
find the slope
AB
m=(10-5)/(5-1)-----> m=5/4
find the slope
CB
m=(15-10)/(9-5)-----> m=5/4
find the slope
DC
m=(20-15)/(13-9)-----> m=5/4
so
<span>the relationship shown is linear
</span>
<span>to model the data with an equation find the equation of a line
</span>with m=5/4 and point A (1,5)
y-y1=m*(x-x1)-----> y-5=(5/4)*(x-1)----> y=1.25x-1.25+5-----> y=1.25x+3.75
the equation to model data is
y=1.25x+3.75
see the attached figure