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2 years ago

First term is is 22 and the common difference is -13

Mathematics

1 answer:

bonufazy [111]2 years ago

8 0

Answer:

$a_{n}$ = 35 - 13n

Step-by-step explanation:

Assuming you require the n th term formula

The n th term of an arithmetic sequence is

$a_{n}$ = a₁ + (n- 1)d

where a₁ is the first term and d the common difference

Here a₁ = 22 and d = - 13, thus

$a_{n}$ = 22 - 13(n - 1) = 22 - 13n + 13 = 35 - 13n

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4x-5≥7 in interval notation

Readme [11.4K]

Answer:

4x - 5 > 7 = [3,∞)

Step-by-step explanation:

Move all terms not containing x to the right side of the inequality.

4x<u> > </u>12

Divide each term by 4 and simplify.

x <u>> </u>3

Convert the inequality to interval notation.

[3,∞)

I hope this helps

if you could, feel free to mark me Brainliest it would be much appreciated :D

3 0

3 years ago

A student used the slope-intercept form to write the

bazaltina [42]

Answer:

it is D: Step 6: student needed to add the b value

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Evaluate : limx→ tan2-sin2x x3

GrogVix [38]

Given:

$\lim _{x\to0}\frac{\tan 2x-\sin 2x}{x^3}$

Solve:

$\lim _{x\to0}\frac{\tan 2x-\sin 2x}{x^3}$

Use l'hopital's rule:

$\begin{gathered} =\lim _{x\to0}\frac{\frac{d}{dx}(-\sin 2x+\tan 2x)}{\frac{d}{dx}(x^3)} \\ =\lim _{x\to0}\frac{-2\cos (2x)+2\tan ^2(2x)+2}{3x^2} \end{gathered}$

Simplify:

$\begin{gathered} =\lim _{x\to0}\frac{-2\cos (2x)+2\tan ^2(2x)+2}{3x^2} \\ =\lim _{x\to0}\frac{2(-\cos (2x)+\tan ^2(2x)+1)}{3x^2} \end{gathered}$

Apply the constant multiple rule:

$\begin{gathered} \lim _{x\to0}cf(x)=c\lim _{x\to0}f(x) \\ \text{With c=}\frac{2}{3} \\ f(x)=\frac{-\cos (2x)+\tan ^2(2x)+1}{x^2} \end{gathered}$ $\begin{gathered} =\frac{2\lim _{x\to0}\frac{-\cos (2x)+\tan ^2(2x)+1}{x^2}}{3} \\ =\frac{2\lim _{x\rightarrow0}\frac{(4\tan ^2(2x)+4)\tan (2x)+2\sin (2x)}{2x}}{3} \end{gathered}$

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