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valentina_108 [34]
3 years ago
9

Help me with this problem

Mathematics
1 answer:
sveta [45]3 years ago
6 0

Answer:

ñññññññññ

Step-by-step explanation:

11/9/2001

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What is log7 1 equal to
igomit [66]

Answer:

1.85

Step-by-step explanation:hope this helps god bless

3 0
3 years ago
Find the slope of the straight line that passes through the following pair of points. (-4,1) and (2,6) ​
loris [4]

Answer:

5/6

Step-by-step explanation:

(6-1)/(2- -4) --> 5/6

7 0
3 years ago
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PLEASE HELP!!! I’LL GIVE BRAINIEST!!!
Vilka [71]

Answer:

  • 8x - 17 = 5x +19

Step-by-step explanation:

Corresponding angles are congruent

thus, 8x - 17 = 5x +19

Hope it helps!

6 0
2 years ago
Read 2 more answers
S t are positive integers
Sindrei [870]

Given:

The expanded form of (x+s)(x-t) is x^2+Kx-40.

Where, s, t, K are positive integers.

To find:

The smallest possible value of K.

Solution:

The expanded form of (x+s)(x-t) is x^2+Kx-40. It means,

(x+s)(x-t)=x^2+Kx-40

x^2-tx+sx-st=x^2+Kx-40

x^2+(s-t)x-st=x^2+Kx-40

On comping both sides, we get

K=s-t              ...(i)

K is a positive integer if s>t.

And

st=40

The factor pairs of 40 are (1,40), (2,20), (4,10), (5,8), (8,5), (10,4), (20,2) and (40,1).

Since s>t, therefore the possible values for (s,t) are (8,5), (10,4), (20,2) and (40,1).

Using (i), find the value of K for each factor pair.

K=8-5=3  

K=10-4=6      

K=20-2=18          

K=40-1=39          

Therefore, the smallest possible value of K is 3.

3 0
3 years ago
∑∞ n=1 (n^2n)/(1+n)^(3n)
kvv77 [185]

Answer:

Given the series,

∑ ∞ n = 1 − 4 ( − 1 / 2 ) n − 1

I think the series is summation from n = 1 to ∞ of -4(-1/2)^(n-1)

So,

∑ − 4 ( − ½ )^(n − 1). From n = 1 to ∞

There are different types of test to show if a series converges or diverges

So, using Ratio test

Lim n → ∞ (a_n+1 / a_n)

Lim n → ∞ (-4(-1/ 2)^(n+1-1) / -4(-1/2)^(n-1))

Lim n → ∞ ((-4(-1/2)^(n) / -4(-1/2)^(n-1))

Lim n → ∞ (-1/2)ⁿ / (-1/2)^(n-1)

Lim n→ ∞ (-1/2)^(n-n+1)

Lim n→ ∞ (-1/2)^1 = -1/2

Since the limit is less than 0, then, the series converge...

Sum to infinity

Using geometric progression formula

S∞ = a / 1 - r

Where

a is first term

r is common ratio

So, first term is

a_1 = -4(-½)^1-1 = -4(-½)^0 = -4 × 1

a_1 = -4

Common ratio r = a_2 / a_1

a_2 = 4(-½)^2-1 = -4(-½)^1 = -4 × -½ = 2

a_2 = 2

Then,

r = a_2 / a_1 = 2 / -4 = -½

S∞ = -4 / 1--½

S∞ = -4 / 1 + ½

S∞ = -4 / 3/2 = -4 × 2 / 3

S∞ = -8 / 3 = -2⅔

The sum to infinity is -2.67 or -2⅔

<h2>Step-by-step explanation: PHEW THAT TOOK A WHILE LOL IM A FAST TYPER</h2>

3 0
3 years ago
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