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Sergeu [11.5K]
2 years ago
12

Anyone has an answer to this question?

Mathematics
1 answer:
mylen [45]2 years ago
8 0

Answer:

1/100 cos100⁰ 100 0.1 tan100⁰

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If you don’t know I don’t answer (NO LINKS ) please helllllppppp
Marina CMI [18]

Answer:

Step-by-step explanation:

1. 6.96 in

2. 2.6 in

3. 8.1 in

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5. 1.89 cm

6. 6 in

6 0
3 years ago
Find the 14th term of the geometric sequence 5,-10,20
tamaranim1 [39]

In this specific case, the <em>initial term</em> (a) is 5 and the <em>common ratio</em> (r) is -2

Henceforth, after determining what a and r are, we use the formula for the <em>nth term</em>. Which is:

a_n = a \times r^{n-1}

Therefore, the 14th term is :

\implies 5 \times  { (- 2)}^{14 - 1}  \\\implies 5 \times  {( - 2)}^{13}  \\ =\boxed { -40,960}

Hope it helps!

7 0
2 years ago
Log2+ log8=2<br><br>how do I do it?​
soldi70 [24.7K]

Answer:

Step-by-step explanation:

log(2)+log(8) = 2

1.20412≠2

False

8 0
3 years ago
Find the volume of the prism:​
Rudiy27

Answer:

240m

Step-by-step explanation:

Volume is length x width x height

7.5m x 4m x 8m: 240m

4 0
3 years ago
g Let G be a not necessarily abelian group with normal subgroups H and K such that H contains K (i.e., K ✂ G, H ✂ G, K ≤ H) and
allsm [11]

Answer:

Lets a,b be elements of G. since G/K is abelian, then there exists k ∈ K such that ab * k = ba (because the class of ab, [ab]_K is equal to [ba]_K, thus ab and ba are equal or you can obtain one from the other by multiplying by an element of K.

Since K is a subgroup of H, then k ∈ H. This means that you can obtain ba from ab by multiplying by an element of H, k. Thus, [ab]_H = [ba]_H . Since a and b were generic elements of H, then H/G is abelian.

4 0
3 years ago
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