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Aleks04 [339]
4 years ago
12

What's 2 3 5 7 11 13 17

Mathematics
1 answer:
krek1111 [17]4 years ago
6 0
They represent the first seven of an infinite series of prime numbers.
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Which expression is equivalent to 24b-34ab ? A- 4b(6-8a) B- 4b(6b-8ab) C- 2b(12-17a) D- 2b(12-17ab)
Montano1993 [528]

Answer:

c.

Step-by-step explanation:

hope this helped

5 0
3 years ago
What is the percentage of 1/4
kati45 [8]
25%

Hers how 100 divided by the denominator multiplied by the numerator will give u an answer in the percentage of any fraction so 100 divided by 4 = 25 multiplied by one = 25
8 0
3 years ago
Read 2 more answers
To the nearest minute, at what time will the hour hand and minute hand be in the same position between 8 and 9 o'clock?
alekssr [168]

Answer:

8:40

Step-by-step explanation:

For the hour hand and the minute hand to be at the same position between 8 and 9 o'clock, then the two of the hands must be at either the 8 hour mark or the 9 hour mark.

If the hour and the minute hand is at point 9, then the time would be 9:45 which is not between 8 and 9 o'clock. But, if the the hour and the minute hand is at point 8, then the time would be 8:40 which is between 8 and 9 o'clock.

3 0
3 years ago
Let Bold u comma Bold v comma Bold w 1 comma Bold w 2 comma and Bold w 3 be vectors in Bold Upper R Superscript n. If u and v ca
Sonja [21]

Answer:

Step-by-step explanation:

given that U, V are two vectors in R^n

These two vectors can be written as a linear combination of 3 vectors

w1, w2, and w3

To prove that  U+V also can be written as a linear combination of these three vectors.

Since U is a linear combination we can write for not all a,b, c equal to 0

U = aw1+bw2+cw3

Similarly for d,e,f not all equal to 0

V= cw1+dw2+ew3

Adding these we have

U+V =(a+d)w1 + (b+e) w2+(c+f)w3

Here all a+d, b+e or c+f cannot be simultaneously 0.

So we get U+V can be written as a linear combination of w1, w2 w3 as follows:

U+W = gw1+hw2+iw3 \\g = a+d\\h = b+e\\i = c+f

Proved

4 0
3 years ago
What is the distance between P(0, 14) and Q(5, -2)?
Sophie [7]
P(x₁ , y₁) and Q(x₂ , y₂) ↔ [ P(0,14)    Q(5, - 2)]
Distance = PQ = √[(x₂ - x₁)² + (y₂ - y₁))²]

PQ = √[(5 - 0)² + (- 2 - 14)²]

PQ =  √[(25 + 256]

PQ =  √281

PQ = 17.76

3 0
4 years ago
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