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2 years ago

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 exactly one solution (often referred to as the unique 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

Cost of each packet will be same as previously as during the offer; $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 :

During offer:

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 = $\frac{21.6}{2400}\times100=\frac{2160}{2400} =\$0.9$

Normal day:

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.

Conclusion:

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