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Elan Coil [88]
3 years ago
15

Please help me out here, I’ve been reposting the same questions and have not seen people not answer mine. :( just please i will

give u brainiest

Mathematics
2 answers:
lilavasa [31]3 years ago
8 0
There are infinite amount of solutions and you can not determine prices of either. So 3 and 6.
anastassius [24]3 years ago
3 0

Answer:

There are infinite solutions because they are the same equation

2D + 1G = 12

and

4D + 2G = 24 (divide by 2) 2D + G = 12

2D + 1G = 12 is the same as 2D + 1G = 12

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An angle measures 48° less than the measure of its complementary angle. What is the measure of each angle?
zepelin [54]

Answer:

42 degrees because both angles have to add to 90 degrees

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
I need help with 2,3 and 6! Please Help!
Georgia [21]
2. If you already know Faulhaber's formula, which says

\displaystyle\sum_{k=1}^nk^2=\dfrac{n(n+1)(2n+1)}6

then it's just a matter of setting n=4. If you don't, then you can prove that it works (via induction), or compute the sum by some other means. Presumably you're not expected to use brute force and just add the squares of 1 through 4.

Just to demonstrate one possible method of verifying the formula, suppose we start from the binomial expansion of (k-1)^3, do some manipulation, then sum over 1\le k\le n:

(k-1)^3=k^3-3k^2+3k-1
\implies k^3-(k-1)^3=3k^2-3k+1
\implies\displaystyle\sum_{k=1}^n(k^3-(k-1)^3)=\sum_{k=1}^n(3k^2-3k+1)

The left side is a telescoping series - several terms in consecutive terms of the series will cancel - and reduces to n^3. For example,

\displaystyle\sum_{k=3}^2(k^3-(k-1)^3)=(1^3-0^3)+(2^3-1^3)+(3^3-2^3)=3^3

Distributing the sum on the right side across each term and pull out constant factors to get

\displaystyle n^3=3\sum_{k=1}^nk^2-3\sum_{k=1}^nk+\sum_{k=1}^n1

If you don't know the formula for \displaystyle\sum_{k=1}^nk=\frac{n(n+1)}2, you can use a similar trick with the binomial expansion (k-1)^2, or a simpler trick due to Gauss, or other methods. I'll assume you know it to save space for the other parts of your question. We then have

\displaystyle n^3=3\sum_{k=1}^nk^2-\frac{3n(n+1)}2+n
\implies\displaystyle\sum_{k=1}^nk^2=\dfrac{n(n+1)(2n+1)}6

and when n=4 we get 30.

3. Each term in the sum is a cube, but the sign changes. Recall that (-1)^n is either 1 if n is even or -1 if n is odd. So we can write

1^3-2^3+3^3-4^3+5^3=\displaystyle\sum_{k=1}^5(-1)^{k-1}k^3

(k+1 as the exponent to -1 also works)

6. If 0\le k\le n, and i=k+1, then we would get 1\le k+1\le n+1\iff1\le i\le n+1. So the sum with respect to i is

\displaystyle\sum_{k=0}^n\frac{k^2}{k+n}=\sum_{i=1}^{n+1}\frac{(i-1)^2}{i+n-1}
4 0
3 years ago
200+5-4+e+9000+100-200
svet-max [94.6K]

Answer:

9103.71828183

Step-by-step explanation:

9103.71828183=200+5-4+e+9000+100-200

7 0
3 years ago
What is the distance between these points (-5,-7) and (1,-2)
Alecsey [184]

Answer:

V61

Step-by-step explanation:

6 0
3 years ago
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The doll was considered a "feminine" toy. Do these data provide convincing evidence that the mean percentage of the time spent p
NikAS [45]

Answer:

yes it does because men comes before women

7 0
3 years ago
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