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svp [43]
3 years ago
6

-5x+13y=-7 5x+4y=24 system of equations

Mathematics
1 answer:
gregori [183]3 years ago
3 0

Answer:

x = 4, y = 1

Step-by-step explanation:

In attachment.

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The product of two consecutive even positive integers is 674 more than the sum of the two integers. Find the integers.
IrinaVladis [17]
2n(2n+2)=2n+2n+2+674\\
4n^2+4n=4n+676\\
4n^2=676\\
n^2=169\\
n=-13 \vee n=13
-13\not\ \textgreater \ 0

n=13\\
2n=26\\
2n+2=28

It's 26 and 28.
7 0
3 years ago
4(x+4) + 2x = 52 need answer
Phoenix [80]

Answer:

x=6

Step-by-step explanation:

3 0
3 years ago
If the length of a rectangle is three times it’s width and it’s parameter is 24cm what is the area
katrin [286]

Answer:

Area = 27 cm²

Step-by-step explanation:

let L be the length of the rectangle

    w the width

    P the perimeter

    A the area

<u>Formulas</u> :

P = 2×(L + w)

A = L × w

<u>We are Given</u> :

L = 3w

P = 2×(L + w)

⇔ 24 = 2×(3w + w)

⇔ 24 = 2×4w

⇔ 24 = 8w

⇔ w = 24/8 = 3

Then

L = 3w = 3×3 = 9

We obtain L = 9 and w = 3.

Then

A = L × w

  = 9 × 3

  = 27

6 0
2 years ago
Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a fam
MariettaO [177]
  • X-(A\cup B)=(X-A)\cap(X-B)

I'll assume the usual definition of set difference, X-A=\{x\in X,x\not\in A\}.

Let x\in X-(A\cup B). Then x\in X and x\not\in(A\cup B). If x\not\in(A\cup B), then x\not\in A and x\not\in B. This means x\in X,x\not\in A and x\in X,x\not\in B, so it follows that x\in(X-A)\cap(X-B). Hence X-(A\cup B)\subset(X-A)\cap(X-B).

Now let x\in(X-A)\cap(X-B). Then x\in X-A and x\in X-B. By definition of set difference, x\in X,x\not\in A and x\in X,x\not\in B. Since x\not A,x\not\in B, we have x\not\in(A\cup B), and so x\in X-(A\cup B). Hence (X-A)\cap(X-B)\subset X-(A\cup B).

The two sets are subsets of one another, so they must be equal.

  • X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices i\in I.

Proof of one direction for example:

Let x\in X-\left(\bigcup\limits_{i\in I}A_i\right). Then x\in X and x\not\in\bigcup\limits_{i\in I}A_i, which in turn means x\not\in A_i for all i\in I. This means x\in X,x\not\in A_{i_1}, and x\in X,x\not\in A_{i_2}, and so on, where \{i_1,i_2,\ldots\}\subset I, for all i\in I. This means x\in X-A_{i_1}, and x\in X-A_{i_2}, and so on, so x\in\bigcap\limits_{i\in I}(X-A_i). Hence X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i).

4 0
3 years ago
Jasmine was expected to get $150 for her bonus, but she only got $100. What was the percent error?
Slav-nsk [51]

Answer:

Jasmine expected to get $150 for her bonus, but she only got $100. What was the percent error?

-----

Amt of error:: $50

----------

% error:: 50/150 = 1/3 = 33 1/3%

7 0
3 years ago
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