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2 years ago

Just need help on this one problem about a triangle

Mathematics

1 answer:

Rainbow [258]2 years ago

4 0

Given:

ΔABC is an equilateral triangle.

AB = 7.5x, BC = 6x + 3 and CA = 10x - 5

To find:

The value of x.

Solution:

In equilateral triangle, all sides are equal.

AB = BC = CA

$7.5x=6x + 3 = 10x - 5$

Take any two sides only:

AB = BC

$7.5 x=6 x+3$

Subtract 6x from both sides.

$7.5 x-6 x=6x+3-6x$

$1.5 x=3$

Divide by 1.5 on both sides.

$$\frac{1.5 x}{1.5}=\frac{3}{1.5}$

$x=2$

The value of x is 2.

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3 years ago

Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:

professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is $\{1,2,3,4\}$ .

a) We need to find a relation R reflexive and transitive that contain the relation $R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}$

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

1R4 and 4R1, then 1 must be related with itself.
4R1 and 1R4, then 4 must be related with itself.
4R1 and 1R2, then 4 must be related with 2.

Therefore $\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\}$ is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

since 1R2, then 2 must be related with 1.
since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

1R2 and 2R1, then 1 must be related with itself.
4R1 and 1R4, then 4 must be related with itself.
2R1 and 1R4, then 2 must be related with 4.
4R1 and 1R2, then 4 must be related with 2.
2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

$\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}$

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

1 must be related with 1,

2 must be related with 2,

3 must be related with 3,

4 must be related with 4

For be symmetric

since 1R2, 2 must be related with 1,
since 1R4, 4 must be related with 1.

For be transitive

Since 4R1 and 1R2, 4 must be related with 2,
since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

$\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}$

5 0

2 years ago

What two numbers multiply to 54 and add up to 7

k0ka [10]

Xy = 54
x + y = 7

$xy = 54$
$\frac{xy}{x} = \frac{54}{x}$
$y = \frac{54}{x}$

$x + y = 7$
$x + \frac{54}{x} = 7$
$\frac{x^{2}}{x} + \frac{54}{x} = 7$
$\frac{x^{2} + 54}{x} = 7$
$x^{2} + 54 = 7x$
$x^{2} - 7x + 54 = 0$
$x = \frac{-(-7) \± \sqrt{(-7)^{2} - 4(1)(54)}}{2(1)}$
$x = \frac{7 \± \sqrt{49 - 216}}{2}$
$x = \frac{7 \± \sqrt{-167}}{2}$
$x = \frac{7 \± i\sqrt{167}}{2}$
$x = \frac{7 \± 12.9i}{2}$
$x = 3.5 \± 6.45i$
$x = 3.5 + 6.45i$ $or$ $3.5 - 6.45i$

x + y = 7
3.5 + 6.45i + y = 7
- (3.5 + 6.45i) - (3.5 + 6.45i)
y = 3.5 - 6.45i
  (x, y) = (3.5 + 6.45i, (3.5 - 6.45i)

  or

  x + y = 7
3.5 - 6.45i + y = 7
- (3.5 - 6.45i)   - (3.5 - 6.45i)
y = 3.5 + 6.45i
(x, y) = (3.5 - 6.45i, 3.5 + 6.45i)

The two numbers that add up to 7 and can multiply to 54 is 3.5 ± 6.45i.

5 0

3 years ago