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balandron [24]
2 years ago
13

The system with one and only one solution set is called a/an system

Mathematics
1 answer:
beks73 [17]2 years ago
6 0

Answer:

Consistent and Independent System

Step-by-step explanation:

A system with at least one solution is a consistent system.  A consistent system is said to be independent if it has <u>exactly one solution</u> (often referred to as the <em>unique</em> solution).  The graphs intersect at exactly one point, which gives an ordered-pair (x, y) as the solution of the system.  

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The jars of peanut butter that I buy cost $7.20 each and normally contain 500 grams. At present, however, each jar contains an e
Alex_Xolod [135]

Answer:

$0.54

Step-by-step explanation:

Given:

The jars of peanut butter that I buy cost $7.20 each and normally contain 500 grams.

At present, however, each jar contains an extra 20% free.

And additionally, the supermarket has a "buy 3 jars get another one free''.

Question asked:

By how much is the cost per 100 grams of peanut butter less than it normally is if I take full advantage of the supermarket's offer today?

Solution:

Cost of a jars of peanut = $7.20

Normally, weight of peanut butter in one packet = 500 gram

At supermarket offer, each jar contains an extra 20% free, means a packet  of peanut butter contains = 500 + 20% of 500                                                                                                            =500+500\times \frac{20}{100} =500+100=600\ gram

And extra benefit = buy 3 jars get another one free, means we will get 4 jars by just giving price of only 3 jars.

That means, total weight of peanut butter you are getting by giving 3 packet's price =  Each jar weigh \times number of jars

                       = 600 \times 4 = 2400 gram

<u>Cost of each packet will be same as previously as during the offer;</u> $7.20

Cost of 3 packet's peanut (2400 gram) =  7.20\times3=\$21.6

Now, we will calculate cost per 100 grams of peanut butter :

<u>During offer:</u>

Cost of 2400 grams of peanut butter = $21.6

Cost of 1 gram of peanut butter = \frac{21.6}{2400}

Cost of 100 gram of peanut butter = \frac{21.6}{2400}\times100=\frac{2160}{2400} =\$0.9

<u>Normal day:</u>

Cost of 500 grams of peanut butter = $7.20

Cost of 1 gram of peanut butter = \frac{7.20}{500}

Cost of 100 gram of peanut butter = \frac{7.20}{500}\times100=\frac{720}{500} =\$1.44

We found that cost of 100 gram of peanut butter in normal day is $1.44 while cost of 100 gram of peanut butter during offer is $0.9 means cost per 100 gram during offer is ($1.44 - $0.9 = )$0.54 less than cost during normal day.

<u>Conclusion:</u>

Therefore, cost per 100 grams of peanut butter is $0.54 less than it is normally if you take full advantage of the supermarket's offer today.

             

7 0
3 years ago
Write the name of the period that has the digits 913 in 913256
solniwko [45]
I don't understand what your asking
8 0
3 years ago
Read 2 more answers
Binomial Expansion/Pascal's triangle. Please help with all of number 5.
Mandarinka [93]
\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}

The rows add up to 1,2,4,8,16, respectively. (Notice they're all powers of 2)

The sum of the numbers in row n is 2^{n-1}.

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When n=1,

(1+x)^1=1+x=\dbinom10+\dbinom11x

so the base case holds. Assume the claim holds for n=k, so that

(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k

Use this to show that it holds for n=k+1.

(1+x)^{k+1}=(1+x)(1+x)^k
(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)
(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}

Notice that

\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}
\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}

So you can write the expansion for n=k+1 as

(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}

and since \dbinom{k+1}0=\dbinom{k+1}{k+1}=1, you have

(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}

and so the claim holds for n=k+1, thus proving the claim overall that

(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n

Setting x=1 gives

(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n

which agrees with the result obtained for part (c).
4 0
3 years ago
You are graphing Square ABCDABCDA, B, C, D in the coordinate plane. The following are three of the vertices of the square: A(4,
Kitty [74]

Answer:

D(4,-3)

Step-by-step explanation:

Given three of the vertices of the square: A(4, -7), B(8, -7),C(8, -3)

Let the coordinate of the fourth vertex be D(x,y).

We know that diagonals of a square are perpendicular bisector. So, the midpoint of both diagonals is the same.

The diagonals are BD and AC

Midpoint of BD = Midpoint of AC

\left(\dfrac{8+x}{2},\dfrac{-7+y}{2}\right) =\left(\dfrac{4+8}{2},\dfrac{-7+(-3)}{2}\right)\\ \left(\dfrac{8+x}{2},\dfrac{y-7}{2}\right) =\left(\dfrac{12}{2},\dfrac{-10}{2}\right)\\ \left(\dfrac{8+x}{2},\dfrac{y-7}{2}\right) =\left(6,-5\right)\\$Therefore$:\\\dfrac{8+x}{2}=6\\8+x=12\\x=12-8\\x=4\\$Similarly$\\\dfrac{y-7}{2}=-5\\y-7=-5*2\\y-7=-10\\y=-10+7=-3

The coordinates of the fourth vertex is D(4,-3)

3 0
3 years ago
8.1+3.8h-5.6h=-7.2 What is h?
mihalych1998 [28]
The answer will be
-1.8h=-7.2-8.1
h=-15.3/-1.8
h=8.5
5 0
3 years ago
Read 2 more answers
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