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german
3 years ago
5

A rectangle has a perimeter of 34 cm and an area of 52 cm2. its length is 5 more than twice its width. write and solve a system

of equations to find the dimensions of the rectangle
Mathematics
1 answer:
Vsevolod [243]3 years ago
3 0
Width=4cm
length=13cm

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agasfer [191]

Answer:

I think its 1234.

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4 0
3 years ago
Read 2 more answers
Please answer thank you
Alchen [17]
-x - 9 \geq 5

You want to get x by itself.

First subtract 9 from both sides of the equation.

-x - 9 \geq 5
    +9                 +9

-x \geq 14

Then divide -1 on both sides.

-x/-1 \geq 14/-1

I'm pretty sure when you divide a negative number when solving inequalities, you switch the sign.

x \leq 5

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3 years ago
Help with this slopeee ASAPPP
anygoal [31]

Answer: 1 over y 2 over x

Step-by-step explanation:

8 0
2 years ago
4 3/10 - ( 2 2/5x+ 5 1/2) = 1/2(-3 3/5x + 1 1/5)
koban [17]

Answer:

x = -223/11 = -20.273

Step-by-step explanation:

Step  1  : Simplify:  11/5

Step 2: Simplify 33/5

Step  3  :

Adding fractions which have a common denominator

Step  4: Simplify: 1/2

Step  5  :  see below

Step  6  :  see below

Pulling out like terms/factors

Step  7 : Simplify: 51/2

Step 8:  Simplify: 22/5

Step 10: Simplify: 43/10

Step  9  :

Calculating the Least Common Multiple

Step  11  :

Adding fractions which have a common denominator

Done.

Hope this helps.

7 0
3 years ago
Let A, B, C and D be sets. Prove that A \ B and C \ D are disjoint if and only if A ∩ C ⊆ B ∪ D
ANEK [815]

Step-by-step explanation:

We have to prove both implications of the affirmation.

1) Let's assume that A \ B and C \ D are disjoint, we have to prove that A ∩ C ⊆ B ∪ D.

We'll prove it by reducing to absurd.

Let's suppose that A ∩ C ⊄ B ∪ D. That means that there is an element x that belongs to A ∩ C but not to B ∪ D.

As x belongs to A ∩ C, x ∈ A and x ∈ C.

As x doesn't belong to B ∪ D, x ∉ B and x ∉ D.

With this, we can say that x ∈ A \ B and x ∈ C \ D.

Therefore, x ∈ (A \ B) ∩ (C \ D), absurd!

It's absurd because we were assuming that A \ B and C \ D were disjoint, therefore their intersection must be empty.

The absurd came from assuming that A ∩ C ⊄ B ∪ D.

That proves that A ∩ C ⊆ B ∪ D.

2) Let's assume that A ∩ C ⊆ B ∪ D, we have to prove that A \ B and C \ D are disjoint (i.e.  A \ B ∩ C \ D is empty)

We'll prove it again by reducing to absurd.

Let's suppose that  A \ B ∩ C \ D is not empty. That means there is an element x that belongs to  A \ B ∩ C \ D. Therefore, x ∈ A \ B and x ∈ C \ D.

As x ∈ A \ B, x belongs to A but x doesn't belong to B.  

As x ∈ C \ D, x belongs to C but x doesn't belong to D.

With this, we can say that x ∈ A ∩ C and x ∉ B ∪ D.

So, there is an element that belongs to A ∩ C but not to B∪D, absurd!

It's absurd because we were assuming that A ∩ C ⊆ B ∪ D, therefore every element of A ∩ C must belong to B ∪ D.

The absurd came from assuming that A \ B ∩ C \ D is not empty.

That proves that A \ B ∩ C \ D is empty, i.e. A \ B and C \ D are disjoint.

7 0
3 years ago
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