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3 years ago

Find the sum of the arithmetic series 11+20+29+...+407

Mathematics

1 answer:

Assoli18 [71]3 years ago

4 0

The Answer is 467
Hops this helps

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Indicate a general rule for the nth term of the sequence a1 = 2 and r = √3.

pantera1 [17]

It is given in the question that,

$a_{1} = 2, r = \sqrt 3$

Since we have the value of r given, so we have to use the formula to find the nth term of the geometric progression, which is

$a_{n} = a(r)^{n-1}$

Substituting the values of a and r, we will get

$a_{n} = 2( \sqrt 3)^{n-1}$

So the correct option is the third option .

3 0

2 years ago

(Picture Included) could someone give me the answers to #6?

Aliun [14]

1/6 I think i might be wrong

7 0

3 years ago

Read 2 more answers

A) Let f(x) = 4x + 2 and g(x) = 2x^2-4. Find the formula for the composition function gof

ollegr [7]

Answer:

g o f = $g(f(x)) = 32x^2 + 32x + 4$

Step-by-step explanation:

Given

$f(x) = 4x +2$

$g(x) =2x^2 - 4$

Required:

Find g o f

This is calculated as:

$gof = g(f(x))$

$g(x) =2x^2 - 4$

So:

$g(f(x)) = 2(4x+2)^2 - 4$

$g(f(x)) = 2(4x+2)(4x+2) - 4$

$g(f(x)) = 2[ 16x^2 + 16x + 4)] - 4$

$g(f(x)) = 32x^2 + 32x + 8 - 4$

$g(f(x)) = 32x^2 + 32x + 4$

5 0

2 years ago

If f(x) = integral of 1/ (sqrt of t^3 +2)dt, then f'1 = ?

mihalych1998 [28]

Differentiating an integral removes the integral.

f(x) = integral of dt/sqrt(t^3 + 2)
f'(x) = 1/sqrt(x^3 + 2)
f'(1) = 1/sqrt(1^3 + 2)
f'(1) = 1/sqrt(3) = sqrt(3)/3.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

4 0

3 years ago

The asseigment that i am doing is kinda hard can i get some help and i am in 7th grade

umka2103 [35]

Answer:

Ok I will help u

Step-by-step explanation:

4 0

2 years ago