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MArishka [77]
3 years ago
15

Roman stands across from a flag pole.

Mathematics
2 answers:
Eva8 [605]3 years ago
3 0
The answer is 13.2 m.
Because we are only given the hypotenuse, adjacent, and theta measurements.
the opposite side is always opposite theta. In this case, theta is 39 degrees. The hypotenuse is always the slant of the triangle, so it is 17 m. And adjacent is obviously the only one left, in this case, it is x.

You have to use sin, cos, or tan for this problem. You would use cos because cos = adjacent /hypotenuse (x/17)

so you would do cos(39) times 17
x = cos(39) * 17
x = .777 * 17
x = 13.21 

And it says round to the nearest tenth, so it is 13.2 m.
Hope this helps.

iris [78.8K]3 years ago
3 0

Answer:13.2

Step-by-step explanation: Took the test can say it is coorect..100%

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help due in 5 min!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!will mark brainliest if correct!!!!!!!!!!!!
ser-zykov [4K]

Answer:

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Step-by-step explanation:

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4 0
3 years ago
Square root of 22801 by long division method
mezya [45]
First step: partition the number you want to square root into a block of 2 digits, starting from the last digit (first diagram)

Second step: As our number is a five-digits, we ends up with 2  28  01. Pick a number that could be squared to get the first partition, 2. This number is 1, since 1×1=1

Third step: Write 1 on the top and on the side, as shown in the second graph

Fourth step: Double the number on the side, which is 1+1=2 and use this number as the first digit for the next multiplier. Meanwhile, subtract 1 from 2 inside the root sign to get 1, then pull the other two digits, 28

Fifth step: We need a value in the boxes that when we multiply together will give a number less than 128. We choose 5 as 25×5=125

Sixth step: Subtract 125 from  128 to give 3, and as the same concept with long division, bring down the 0 and the 1. So we have 301

Seventh step: Add 5 to the multiplier on the left, so 20+5=30, which we will use on the side as the hundred and ten digits.

Final step: Find a number to fit in the boxes. We choose 1 since 301×1=301

And hence the square root of 228801 is 151

3 0
3 years ago
23. A population of mice grows according to the formula A(t) = 300e0.087t where
dem82 [27]

Option B

The amount of mice when t = 0 is 300

<h3><u>Solution:</u></h3>

Given that  population of mice grows according to the formula:

A(t) = 300e^{0.087t}

where  t is measured in months

<em><u>To find: amount of mice when t = 0</u></em>

So we have to substitute t = 0 in given formula and find value of A(t)

A(t) = 300e^{0.087t}

Substitute t = 0

A(t) = 300e^{0.087 \times 0}\\\\A(t) = 300e^{0}

\text{ We know that } e^0 = 1

So we get,

A(t) = 300 \times 1 = 300

Thus the amount of mice when t = 0 is 300

7 0
4 years ago
Michael rented a truck for one day. There was a base fee of $20.95, and there was an additional charge of 98 cents for each mile
jenyasd209 [6]

Answer: He drove 139 miles.

Step-by-step explanation:

0.98x + 20.95= 157.17  where x is the number of miles driven.

             -20.95 =-20.95

0.98x = 136.22

x= 139

3 0
4 years ago
If c is the line segment connecting (x1,y1) to (x2,y2), show that the line integral of xdy-ydx=x1y2-x2y1......this is in a chapt
MatroZZZ [7]
Green's theorem doesn't really apply here. GT relates the line integral over some *closed* connected contour that bounds some region (like a circular path that serves as the boundary to a disk). A line segment doesn't form a region since it's completely one-dimensional.

At any rate, we can still compute the line integral just fine. It's just that GT is irrelevant.

We parameterize the line segment by

\mathbf r(t)=\langle x_1,y_1\rangle(1-t)+\langle x_2,y_2\rangle t
\implies\mathbf r(t)=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle

with 0\le t\le1. Then we find the differential:

\mathbf r(t)\equiv\langle x,y\rangle=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle
\implies\mathrm d\mathbf r\equiv\langle\mathrm dx,\mathrm dy\rangle=\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt

with 0\le t\le1.

Here, the line integral is

\displaystyle\int_{\mathcal C}x\,\mathrm dy-y\,\mathrm dx=\int_{\mathcal C}\langle-y,x\rangle\cdot\langle\mathrm dx,\mathrm dy\rangle
=\displaystyle\int_{t=0}^{t=1}\langle-y_1-(y_2-y_1)t,x_1+(x_2-x_1)t\rangle\cdot\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt
=\displaystyle\int_{t=0}^{t=1}(x_1y_2-x_2y_1)\,\mathrm dt
=(x_1y_2-x_2y_1)\displaystyle\int_{t=0}^{t=1}\,\mathrm dt
=x_1y_2-x_2y_1

as required.
4 0
4 years ago
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