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3 years ago

Pls help I’ll give you 35 points

Mathematics

2 answers:

Zarrin [17]3 years ago

6 0

Answer:

next two number10,20,40

Sindrei [870]3 years ago

4 0

Answer:

6.25, 12.5

Step-by-step explanation:

The sequence goes divide by 4 then multiply by 2

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Ahmed is working at a restaurant . His boss pays him $15.50 per tour and promises a raise of $1.25 per hour every 6 months. Whic

aksik [14]

Answer:

Step-by-step explanation:

you have to make chart. You have to add 1.25 to your total every 6 months.

Start $15.50
6 months $15.75
12 months $18.00
18 months. $19.25

5 0

3 years ago

WILL GIVE BRAINLIEST! Find k so that (5, k) is equidistant from (–1, 2) and (3, 0).

pochemuha

Answer:

Step-by-step explanation:

So first I created 2 distance forumlas and set them equal to each other

$\sqrt{(5+1)^2+(k-2)^2} =\sqrt{(5-3)^2+(k)^2}$

then I simplified them

$(5+1)^2+(k-2)^2 =(5-3)^2+(k)^2\\$

$36 + k^2 - 4k +4 = k^2 + 4$

$-4k + 36 = 0\\$

$4k = 36$

$k = 9$

when k = 9, the two distances are equal

7 0

4 years ago

Can someone help me with this i have been trying for hours and i actually got the wrong answer so can you help me understand by

kap26 [50]

Answer:

Step-by-step explanation:

We are given that FGH is a triangle, which means that...

Angle F + Angle G + Angle H = 180

Start by substituting the given angle measurements into the equation:

4x+2+13x-7+3x+5=180

Combine like terms

20x=180

Divide both sides by 20 to isolate x

x=9

Plug 9 back in for x to solve for each angle

Angle F = 4(9)+2=36+2=38

Angle G = 13(9)-7=117-7=110

Angle H = 3(9)+5=27+5=32

7 0

2 years ago

Read 2 more answers

Find maclaurin series

Mumz [18]

Recall the Maclaurin expansion for cos(x), valid for all real x :

$\displaystyle \cos(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}$

Then replacing x with √5 x (I'm assuming you mean √5 times x, and not √(5x)) gives

$\displaystyle \cos\left(\sqrt 5\,x\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt5\,x\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^{2n}}{(2n)!}$

The first 3 terms of the series are

$\cos\left(\sqrt5\,x\right) \approx 1 - \dfrac{5x^2}2 + \dfrac{25x^4}{24}$

and the general n-th term is as shown in the series.

In case you did mean cos(√(5x)), we would instead end up with

$\displaystyle \cos\left(\sqrt{5x}\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt{5x}\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^n}{(2n)!}$

which amounts to replacing the x with √x in the expansion of cos(√5 x) :

$\cos\left(\sqrt{5x}\right) \approx 1 - \dfrac{5x}2 + \dfrac{25x^2}{24}$

7 0

2 years ago

Samantha and Jerrod were at the park, sitting on a park bench together. They decided to look at a fountain on their way home. Je

PilotLPTM [1.2K]

Answer:

Step-by-step explanation:

hdhf

3 0

3 years ago