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3 years ago

Plzz hurry ill give brainliest

Mathematics

2 answers:

Lunna [17]3 years ago

7 0

Answer:

Yeet

Step-by-step explanation:

abruzzese [7]3 years ago

4 0

Answer:

18c-s

Step-by-step explanation:

Spike. dont. Trust. It

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Pleaseee i rlly need this one rn!!

bija089 [108]

Answer:

its 20 inches

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

What is the domain of the cubic function -(2x+3)^3 -1

Mazyrski [523]

Answer:

(-infinite,+infinite)

7 0

3 years ago

Can someone help me and explain? I will mark brainlest. ♡

Sedbober [7]

Answer:

$p'(4) = -3$

$q'(8) = \frac{1}{4}\\$

Step-by-step explanation:

For p'(4):

$p(x) = f(x)g(x) \\ p'(x) = \frac{d}{dx}(f(x)g(x)) \\ p'(x) = f'(x)g(x) +f(x)g'(x)$

$p'(4) = f'(4)g(4) + f(4)g'(4) \\ p'(4) = (-1)(3) +(7)(0) \\ p'(4) = -3$

For q'(8):

$q(x) = \frac{f(x)}{g(x)} \\ q'(x)= \frac{d}{dx}(\frac{f(x)}{g(x)}) \\ q'(x) = \frac{f'(x)g(x) -f(x)g'(x)}{{g(x)}^2}$

$q'(8) = \frac{f'(8)g(8) -f(8)g'(8)}{{g(8)}^2} \\ q'(8) = \frac{(2)(2) -(6)(\frac{1}{2})}{{2}^2} \\ q'(8) = \frac{4 -3}{4} \\ q'(8) = \frac{1}{4}$

4 0

3 years ago

Read 2 more answers

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

How to solve the function 4y+x=12

Sidana [21]

First you rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation

Second you solve it by using the formula of a straight line drawn on Cartesian coordinate system in which “y” is the vet risk axis and “x” the horizontal axis.

7 0

2 years ago