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Likurg_2 [28]
3 years ago
12

Find the missing value to the nearest hundredth.. . tan ___ = 73

Mathematics
2 answers:
Natalka [10]3 years ago
7 0

<span>The missing value to the nearest hundredth for tan___ = 73 is </span>89.22. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

Alex Ar [27]3 years ago
5 0

Answer:

The missing value is 89.22\°

Step-by-step explanation:

Let

x------> the missing angle

we have that

tan(x\°)=73

Using a calculator

x\°=arctan(73)=89.22\°

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Expanded form - (0×1+9×1/10+1×1/100+5×1/1000) miles per minute.

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Enter the expression using exponents. 11 × 11 × 11 × 11 × 11 =
Alla [95]

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the answer should be 11^{5} or 11^5

Explanation:

Since there are five 11 in the equation, the answer should be 11^{5} or 11^5

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6 0
3 years ago
The sequence$$1,2,1,2,2,1,2,2,2,1,2,2,2,2,1,2,2,2,2,2,1,2,\dots$$consists of $1$'s separated by blocks of $2$'s with $n$ $2$'s i
kicyunya [14]

Consider the lengths of consecutive 1-2 blocks.

block 1 - 1, 2 - length 2

block 2 - 1, 2, 2 - length 3

block 3 - 1, 2, 2, 2 - length 4

block 4 - 1, 2, 2, 2, 2 - length 5

and so on.


Recall the formula for the sum of consecutive positive integers,

\displaystyle \sum_{i=1}^j i = 1 + 2 + 3 + \cdots + j = \frac{j(j+1)}2 \implies \sum_{i=2}^j = \frac{j(j+1) - 2}2

Now,

1234 = \dfrac{j(j+1)-2}2 \implies 2470 = j(j+1) \implies j\approx49.2016

which means that the 1234th term in the sequence occurs somewhere about 1/5 of the way through the 49th 1-2 block.

In the first 48 blocks, the sequence contains 48 copies of 1 and 1 + 2 + 3 + ... + 47 copies of 2, hence they make up a total of

\displaystyle \sum_{i=1}^48 1 + \sum_{i=1}^{48} i = 48+\frac{48(48+1)}2 = 1224

numbers, and their sum is

\displaystyle \sum_{i=1}^{48} 1 + \sum_{i=1}^{48} 2i = 48 + 48(48+1) = 48\times50 = 2400

This leaves us with the contribution of the first 10 terms in the 49th block, which consist of one 1 and nine 2s with a sum of 1+9\times2=19.

So, the sum of the first 1234 terms in the sequence is 2419.

8 0
2 years ago
Answer:
emmainna [20.7K]

Let the numbers be n, n+2, n+4

Sum equals too= 13+2(n+4), which is 2n+21

a) Equation--> n+n+2+n+4= 2n+21

b) Solution--> 3n+6= 2n+21

                      => n= 15

c) Second number--> 17 (15+2)

   Third number--> 19 (15+4)

d) 15+15+2+15+4=30+21

   => 51= 51

So, the equation is true.

And pls mark me brainliesttt :)))

8 0
2 years ago
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