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DanielleElmas [232]
3 years ago
9

Which sequence matches the recursive formula?

Mathematics
2 answers:
Ostrovityanka [42]3 years ago
8 0
A1 = 2
a2 = (a1)^2 + 3 = 2^2 + 3 = 4 + 3 = 7
a3 = (a2)^2 + 3 = 7^2 + 3 = 49 + 3 = 52
a4 = (a3)^2 + 3 = 52^2 + 3 = 2,704 + 3 = 2,707

Option C is the correct answer.
marshall27 [118]3 years ago
8 0

Answer:

The solution is 2, 7, 52, 2707, .... To find each term in the sequence, you square the previous term and add 3. This sequence is represented by the recursive formula an = (an-1)2 + 3, where a1 = 2.

So It's C.

Step-by-step explanation:

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When a factory operates from 6 AM to 6 PM, its total fuel consumption varies according to the formula f(t)=0.9t^3−0.6t^0.5+10, w
Nataly [62]

Step-by-step explanation:

the answer is in the image above

7 0
3 years ago
Find the second of two consecutive integers if the second is 13 less than twice the first.
emmasim [6.3K]
What is the range in which the integers can be in? I cannot fully answer this question. But I think there could be more than one answer. One answer would be the first is 36 and the second is 66.

Hope this helps!
Can u plz mark me as brainliest? I really need it!
4 0
3 years ago
If 3x-13=8, find the value of -4?
sukhopar [10]
If you're looking for x then it's 7. The value of -4 is -4.
7 0
3 years ago
3.14=7x-2y
eimsori [14]

3.14=7x-2y

move 7x to the other side

sign changes from +7x to -7x

3.14-7x=7x-7x-2y

3.14-7x=-2y

Divide both sides by -2

3.14/-2-7x/-2=-2y/-2

3.14/-2=-1.57

-7x/-2=7x/2

Answer:

y=-1.57+7x/2

5 0
4 years ago
* Let S = Span {(2,-1, 1), (3, 1, 1), (1, 2, 0)}. (i) Calculate the dimension of S.
Sholpan [36]

The span of 3 vectors can have dimension at most 3, so 9 is certainly not correct.

Check whether the 3 vectors are linearly independent. If they are not, then there is some choice of scalars c_1,c_2,c_3 (not all zero) such that

c_1 (2,-1,1) + c_2 (3,1,1) + c_3 (1,2,0) = (0,0,0)

which leads to the system of linear equations,

\begin{cases} 2c_1 + 3c_2 + c_3 = 0 \\ -c_1 + c_2 + 2c_3 = 0 \\ c_1 + c_2 = 0 \end{cases}

From the third equation, we have c_1=-c_2, and substituting this into the second equation gives

-c_1 + c_2 + 2c_3 = 2c_2 + 2c_3 = 0 \implies c_2 + c_3 = 0 \implies c_2 = -c_3

and in turn, c_1=c_3. Substituting these into the first equation gives

2c_1 + 3c_2 + c_3 = 2c_3 - 3c_3 + c_3 = 0 \implies 0=0

which tells us that any value of c_3 will work. If c_3 = t, then c_1=t and c_2 = -t. Therefore the 3 vectors are not linearly independent, so their span cannot have dimension 3.

Repeating the calculations above while taking only 2 of the given vectors at a time, we see that they are pairwise linearly independent, so the span of each pair has dimension 2. This means the span of all 3 vectors taken at once must be 2.

3 0
2 years ago
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