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3 years ago

Can someone explain how we solve this problem

Mathematics

2 answers:

Virty [35]3 years ago

5 0

X + 5 - 1/x = x - 7 + 1/x
Bring x + 5 - 1/x together using the common denominator x. Bring x - 7 + 1/x together using the common denominator x:
(x^2 + 5 x - 1)/x = (x^2 - 7 x + 1)/x
Multiply both sides by x:
x^2 + 5 x - 1 = x^2 - 7 x + 1
Subtract x^2 - 7 x + 1 from both sides:
12 x - 2 = 0
Factor constant terms from the left hand side:
2 (6 x - 1) = 0
Divide both sides by 2:
6 x - 1 = 0
Add 1 to both sides:
6 x = 1
Divide both sides by 6:
Answer: x = 1/6

riadik2000 [5.3K]3 years ago

4 0

It is look so x=? so if it was 2 then it would be
2-1 = 2+1
_______ which is wrong so that's how to do it which is x?
2+5 = 2-7

the answer is 1/7 so ya its D.1/7 :) hope i helped

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26. Define a relation ∼ ∼ on R 2 R2 by stating that ( a , b ) ∼ ( c , d ) (a,b)∼(c,d) if and only if a 2 + b 2 ≤ c 2 + d 2 . a2+

Tresset [83]

Answer:

~ is reflexive.

~ is asymmetric.

~ is transitive.

Step-by-step explanation:

~ is reflexive:

i.e., to prove $\forall (a, b) \in \mathbb{R}^2$ , $(a, b) R(a, b)$ .

That is, every element in the domain is related to itself.

The given relation is $\sim: (a,b) \sim (c, d) \iff a^2 + b^2 \leq c^2 + d^2$

Reflexive:

$(a, b) \sim (a, b)$ since $a^2 + b^2 = a^2 + b^2$

This is true for any pair of numbers in $\mathbb{R}^2$ . So, $\sim$ is reflexive.

Symmetry:

$\sim$ is symmetry iff whenever $(a, b) \sim (c, d)$ then $(c, d) \sim (a, b)$ .

Consider the following counter - example.

Let (a, b) = (2, 3) and (c, d) = (6, 3)

$a^2 + b^2 = 2^2 + 3^2 = 4 + 9 = 13$

$c^2 + d^2 = 6^2 + 3^2 = 36 + 9 = 42$

Hence, $(a, b) \sim (c, d)$ since $a^2 + b^2 \leq c^2 + d^2$

Note that $c^2 + d^2 \nleq a^2 + b^2$

Hence, the given relation is not symmetric.

Transitive:

$\sim$ is transitive iff whenever $(a, b) \sim (c, d) \hspace{2mm} \& \hspace{2mm} (c, d) \sim (e, f)$ then $(a, b) \sim (e, f)$

To prove transitivity let us assume $(a, b) \sim (c, d)$ and $(c, d) \sim (e, f)$ .

We have to show $(a, b) \sim (e, f)$

Since $(a, b) \sim (c, d)$ we have: $a^2 + b^2 \leq c^2 + d^2$

Since $(c, d) \sim (e, f)$ we have: $c^2 + d^2 \leq e^2 + f^2$

Combining both the inequalities we get:

$a^2 + b^2 \leq c^2 + d^2 \leq e^2 + f^2$

Therefore, we get: $a^2 + b^2 \leq e^2 + f^2$

Therefore, $\sim$ is transitive.

Hence, proved.

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3 years ago

Cual es la formula de aceleracion

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3 years ago

If we apply the Pythagorean Theorem to the Egyptian knots we can see that 3^2 + 4^2 = 5^23 2 +4 2 =5 2 . Which of the sets of th

pantera1 [17]

Answer:

B) { 51, 149, 140 }

C) { 16, 63, 65 }

Step-by-step explanation:

We need to choose Pythagoras theorem formula

$c^2=a^2+b^2$

now, we can verify each options

option-A:

a=1

b=1

c=1

now, we can verify formula

$1^2=1^2+1^2$

$1=2$

We can see that left side is not equal to right side

so, this is FALSE

option-B

a=51

b=140

c=149

now, we can verify formula

$149^2=51^2+140^2$

$22201=22201$

We can see that left side is equal to right side

so, this is TRUE

option-C

a=16

b=63

c=65

now, we can verify formula

$65^2=16^2+63^2$

$4225=4225$

We can see that left side is equal to right side

so, this is TRUE

option-D:

a=6

b=8

c=11

now, we can verify formula

$11^2=6^2+8^2$

$121=100$

We can see that left side is not equal to right side

so, this is FALSE

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3 years ago

Read 2 more answers

Im toooo bad at math plz help

balandron [24]

Answer:

-3/22

Step-by-step explanation:

Mulitiply straight across. Num * num/ denom * denom.

Now you have -6/44. Divide both by 2 to simplify into simipliest form.

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2 years ago

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If the zeros of a quadratic function, F, are -2 and 4, what is the equation of the axis of symmetry of F

vagabundo [1.1K]

Answer:

x=1

Step-by-step explanation:

if the zeroes are (-2,0) and (4,0), then the axis of symetry is going to be right in the middle of those two, with the graph reflected equally on either side.

so you get:

(4-2)÷2=1

therefore the equation of the axis of symetry is x=1

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2 years ago