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ololo11 [35]
3 years ago
6

2/5 divided by 3 and 1/10

Mathematics
2 answers:
Jlenok [28]3 years ago
5 0
Simple but the answer haha. Hope this helps. Good luck!!

Licemer1 [7]3 years ago
3 0
2/5ths divides by 3 and 1/10 is 0.129
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Lines a and b are parallel. Find the measure of each angle. Angle 8 is 113 degrees
pentagon [3]

Answer:

angles 1 4 and 5 are all equal to 113 and angles 2 3 6 and 7 equal 67

3 0
3 years ago
Can someone help me with my algebra class?
Lilit [14]
Im stuck on algebra too
3 0
3 years ago
How do the values in Pascal’s triangle connect to the coefficients?
damaskus [11]

Explanation:

Each row in Pascal's triangle is a listing of the values of nCk = n!/(k!(n-k)!) for some fixed n and k in the range 0 to n. nCk is <em>the number of combinations of n things taken k at a time</em>.

If you consider what happens when you multiply out the product (a +b)^n, you can see where the coefficients nCk come from. For example, consider the cube ...

  (a +b)^3 = (a +b)(a +b)(a +b)

The highest-degree "a" term will be a^3, the result of multiplying together the first terms of each of the binomials.

The term a^b will have a coefficient that reflects the sum of all the ways you can get a^b by multiplying different combinations of the terms. Here they are ...

  • (a +_)(a +_)(_ +b) = a·a·b = a^2b
  • (a +_)(_ +b)(a +_) = a·b·a = a^2b
  • (_ +b)(a +_)(a +_) = b·a·a = a^2b

Adding these three products together gives 3a^2b, the second term of the expansion.

For this cubic, the third term of the expansion is the sum of the ways you can get ab^2. It is essentially what is shown above, but with "a" and "b" swapped. Hence, there are 3 combinations, and the total is 3ab^2.

Of course, there is only one way to get b^3.

So the expansion of the cube (a+b)^3 is ...

  (a +b)^3 = a^3 + 3a^2b +3ab^2 +b^3 . . . . . with coefficients 1, 3, 3, 1 matching the 4th row of Pascal's triangle.

__

In short, the values in Pascal's triangle are the values of the number of combinations of n things taken k at a time. The coefficients of a binomial expansion are also the number of combinations of n things taken k at a time. Each term of the expansion of (a+b)^n is of the form (nCk)·a^(n-k)·b^k for k =0 to n.

6 0
3 years ago
Find the discriminat of each quadratic equation then state the number and type of solutions
Julli [10]
Hello : 
the discriminat of each quadratic equation : ax²+bx+c=0 ....(a <span>≠ 0) is :
</span><span>Δ = b² -4ac
1 )  </span>Δ > 0  the equation has two reals solutions : x =  (-b±√Δ)/2a
2 ) Δ = 0 : one solution : x = -b/2a
3 ) Δ <span>< 0 : no reals solutions</span>
8 0
3 years ago
Car park algebra question
Svetllana [295]
You did south, north, correctly

but last one

twice north is 2 times (x+48) or 2x+96


so

s=x
n=x+48
c=4x
s+c>2x+96
subsitute s for x and 4x for c
x+4x>2x+96
5x>2x+96
minus 2x both sides
3x>96
divide both sides by 3
x>32
it cannot be equal to 32 so
the minimum value is 33 spaces in south car park
5 0
3 years ago
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