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Volgvan
3 years ago
14

Solve for x, given: x2−4x+29=0

Mathematics
2 answers:
SVEN [57.7K]3 years ago
5 0

Answer:

X= 5i-2, x=5i-2

Step-by-step explanation:

lianna [129]3 years ago
3 0

Step-by-step explanation:

{x}^{2}  - 4x + 29 = 0 \\ x =  \frac{ - b± \sqrt{ {b}^{2} - 4ac } }{2a}  \\ x =  \frac{4± \sqrt{ - 100} }{2}  \\ x =  \frac{4±10i}{2}  \\ x = 2±5i \\ x = 2 + 5i \:  \: or \:  \: x = 2 - 5i

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LaTanya leaves her house at 12:30pm and bikes at 12 mi/h to marta’s House she stays at marta’s House for 90mij both girls walk b
babymother [125]

Answer:

2.9 mi

Step-by-step explanation:

The time difference t between 12.30pm and 3.30pm is 3h.

Let the distance to Marat's house be s.

The equation to calculate velocity v is given by:

v = distance / time.

Now you can write an equation for the time difference t and use the two velocities and the 90 min(= 1.5h) she stays:

3 = s/12 + 1.5 + s/2.3

Solve this equation for s:

(1/12 + 10/23)* s = 1.5

s = 2.9

7 0
3 years ago
Determine formula of the nth term 2, 6, 12 20 30,42​
nalin [4]

Check the forward differences of the sequence.

If \{a_n\} = \{2,6,12,20,30,42,\ldots\}, then let \{b_n\} be the sequence of first-order differences of \{a_n\}. That is, for n ≥ 1,

b_n = a_{n+1} - a_n

so that \{b_n\} = \{4, 6, 8, 10, 12, \ldots\}.

Let \{c_n\} be the sequence of differences of \{b_n\},

c_n = b_{n+1} - b_n

and we see that this is a constant sequence, \{c_n\} = \{2, 2, 2, 2, \ldots\}. In other words, \{b_n\} is an arithmetic sequence with common difference between terms of 2. That is,

2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2

and we can solve for b_n in terms of b_1=4:

b_{n+1} = b_n + 2

b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2

b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2

and so on down to

b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)

We solve for a_n in the same way.

2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)

Then

a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)

a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))

a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))

and so on down to

a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2

\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}

6 0
2 years ago
Find the equation of the horizontal line passing through the point (-7 , -3)
Marta_Voda [28]

y = - 3

the equation of a horizontal line, parallel to the x- axis is y = c where c is the value of the y- coordinate the line passes through

Line passes through ( - 7, - 3) → y- coordinate is - 3

equation of line is y = - 3


4 0
3 years ago
Can someone please help me with this question?
dimulka [17.4K]
Answer: Hahahhahahaha


Explanation:

Ihwhwbwbajbwbausbw ahjw
6 0
3 years ago
Line WX is parallel to line YZ. if m
Nikitich [7]

X = 84

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5 0
3 years ago
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