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3 years ago

ASAP! I WILL GIVE BRAINLIEST! LOVE YOU FOREVER!

Mathematics

2 answers:

Readme [11.4K]3 years ago

8 0

Answer:

ODIN THOR 9 WORLDS VANIR VALHALLA FREY FREYA LOKI

Step-by-step explanation:

iren2701 [21]3 years ago

6 0

Answer:

The 2 Clans are Æsir and Vanir.

The gods are Baldur, Bil, Borr, Bragi, Buri, Byggvir, Dellingr, Forseti, Freyr, Henir, Hermod, Lodurr, Mani, Meili, Njord, Ve, Vili, Thor, Odin, Frigg, Loki, Hod, and Tyr.

there are 9 world's: Niflheim, Muspelheim, Asgard, Midgard, Jotunheim, Vanaheim, Alfheim, Svartalfheim, and Helheim.

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Dan has some marbles

Andreyy89

A. Frank has 24 marbles. I found this because Dan has 15 and total is 63 so subtract 63-15= 48. 48 divided by 2 people, which are Ellie ans Dan is 24. Dan has 24 marbles.

3 0

3 years ago

NO LINKS!!! 14. Consider the table

iVinArrow [24]

Answer:

(a) rate of change = 8

(b) Function: y = 8x

Range: 0 ≤ y ≤ 16

Explanation:

Looks like a inverse variation sequence.

Inverse Variation formula: y = k(x)

Take two points: (1, 8), (2, 16)

Find the value of k which is constant also considered as rate of change.

Insert values:

8 = k(1)

k = 8

===========

16 = k(2)

k = 8

Equation: y = 8x

Domain lies in the x axis, Range lies in the y axis.

Here domain: 0 ≤ x ≤ 2

Here range: 0 ≤ y ≤ 16

5 0

1 year ago

Read 2 more answers

Find m∠ABD and m∠CBD given m∠ABC = 111∘.

sineoko [7]

Answer:

m∠ABD = 88º

m∠CBD = 23º

Step-by-step explanation:

(-10x + 58) + (6x + 41) = 111

Combine like terms

-4x + 99 = 111

Subtract 99 from both sides

-4x = 12

Divide both sides by -4

x = -3

------------------------

m∠ABD = -10x + 58

m∠ABD = -10(-3) + 58

m∠ABD = = 30 + 58

m∠ABD = 88º

m∠CBD = 6x + 41

m∠CBD = 6(-3) + 41

m∠CBD = -18 + 41

m∠CBD = 23º

7 0

2 years ago

Read 2 more answers

[(2 times 10) divided by 1/3] + 8 =

kondaur [170]

14.6666666667

Step-by-step explanation:

2•10=20/1/3=6.66666666667+8 is that but you can write it like 14.6 with _ that on top

3 0

2 years ago

If f (n)(0) = (n + 1)! for n = 0, 1, 2, , find the taylor series at a=0 for f.

Pie

Given that $f^{(n)}(0)=(n+1)!$ , we have for $f(x)$ the Taylor series expansion about 0 as

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty(n+1)x^n$

Replace $n+1$ with $n$ , so that the series is equivalent to

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}$

and notice that

$\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty nx^{n-1}$

Recall that for $|x|$ , we have

$\displaystyle\sum_{n=0}^\infty x^n=\frac1{1-x}$

which means

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}=\frac{\mathrm d}{\mathrm dx}\frac1{1-x}$
$\implies f(x)=\dfrac1{(1-x)^2}$

5 0

3 years ago