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Nady [450]
3 years ago
13

The price of a flight was increased by 3% to £650. What was the price before the increase? Give your answer to the nearest penny

.
Mathematics
1 answer:
kramer3 years ago
4 0

Answer:

269

Step-by-step explanation:

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Simplify the following completely, show all work. √-45
scoray [572]

Answer:

3\sqrt{5}i

Step-by-step explanation:

\sqrt{-45}

\sqrt{-9*5}

\sqrt{-9}\sqrt{5}

3i\sqrt{5}

3\sqrt{5}i

7 0
2 years ago
How many different three-letter initials with none of the letters repeated can people have?
viva [34]
This is a probablity question

there are 3 slots, 26 letters of the alphabet

so in the first slot, we have 26 possible letters
2nd slot, we have 25 possible letters (we used one for the first slot)
3rd slot, we have 24 possible letters, (we used 2 for first and 2nd slot)

so in total, we have 26*25*24 or 15600 different 3 letter initials with no repeated letters
3 0
3 years ago
Larissa is writing a coordinate proof to show that the diagonals of a square are perpendicular to each other. She starts by assi
Llana [10]

The slope of KH is 1

The slope of GJ is -1

The product of the slopes of the diagonals is -1

Therefore, KH is perpendicular to GJ

4 0
3 years ago
Pls answer this for meeee​
Roman55 [17]

Answer:

7 hamsters at 5 pesos each and 13 hamsters at 8 pesos each

Step-by-step explanation:

5x + 8y = 139

x + y = 20

x = 20 - y

5x + 8y = 139

5(20 - y) + 8y = 139

100 - 5y + 8y = 139

100 + 3y = 139

3y = 139 - 100

3y = 39

3y/3 = 39/3

y = 13

x + y = 20

x + 13 = 20

x = 20 - 13

x = 7

7 0
3 years ago
Which of the following has a solution set of {x | x = 0}?
notka56 [123]

Answer:

(b)  (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) =  {x | x = 0}

Step-by-step explanation:

Here, the given expression is : {x | x = 0}

So, the ONLY element in the given set = {0}

Now, take each option and solve the given expression:

(a)  x + 1 < -1

Adding -1 BOTH sides, we get:

x + 1 -1 < -1  -1

or, x < - 2 ⇒ x = { -∞ , .... , -4,-3}

Also,   x + 1 < 1

Adding -1 BOTH sides, we get:

x + 1 -1 < 1  -1

or, x <0 ⇒ x = { -∞ , .... , -4,-3,-2,-1}

So, (x + 1 < -1) ∩ (x + 1 < 1) =  { -∞ , .... , -4,-3}∩ { -∞ , .... , -4,-3,-2,-1}  

=  { -∞ , .... , -4,-3}

⇒ (x + 1 < -1) ∩ (x + 1 < 1) ≠ {0}

Similarly, solving

(b) (x + 1 ≤ 1) ∩ (x + 1 ≥ 1)

(x + 1 ≤ 1) =  x≤ 0 =   { -∞ , .... , -4,-3,-2,-1, 0}

(x + 1 ≥ 1) =  x ≥ 0 =  {0,1,2,3,... ∞}

⇒ (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) = {0}

(b)(x + 1 < 1) ∩ (x + 1 > 1)

(x + 1 < 1) =  x <  0 =   { -∞ , .... , -4,-3,-2,-1}

(x + 1 > 1) =  x  > 0 =  { 1,2,3,... ∞}

⇒ (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) = Ф

Hence,  (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) =  {x | x = 0}

8 0
4 years ago
Read 2 more answers
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