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3 years ago

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

2 answers:

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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A car travels at a speed of 0.915 miles per minute. Write the speed in expanded<br> form.

lisabon 2012 [21]

Answer:

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

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2 years ago

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AlexFokin [52]

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3 years ago

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Enter the expression using exponents. 11 × 11 × 11 × 11 × 11 =

Alla [95]

Answer:

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

You shouldn't write 161051 as the answer because it is NOT a exponent.

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