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Leya [2.2K]
3 years ago
11

Choose the polynomial that is written in standard form.

Mathematics
1 answer:
eduard3 years ago
8 0

Answer:

B OR the second one

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The slope of the line below is 2. Which of the following is the point-slope form of the line? the x is 1 and the y is -1
sukhopar [10]
The point slope form would be y+1=m(x-1)


5 0
3 years ago
Read 2 more answers
Please help due tomorrow
icang [17]

Answer:

Here's What you do:

  1. Draw a line that comes straight through the middle
  2. It has to create 4 - 90° angles (use a protractor to measure)

There's a picture attached on how a perpendicular lines should look like.

Step-by-step explanation:

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

4 0
3 years ago
Ten samples of coal from a northern appalachian source had an average mercury content of 0.242 ppm with a standard deviation of
Dovator [93]
The confidence interval is from 0.2191 to 0.2649, given by

0.242\pm0.0229.

The z-score we need is found by first subtracting the confidence level from 1:
1 - 0.95 = 0.05

Divide that by 2:
0.05/2 = 0.025

Subtract that from 1:
1 - 0.025 = 0.975

Look this up in the middle of a z-table (http://www.z-table.com) and we see that the z-score is 1.96.

Now our confidence interval is given by
\overline{x} \pm z\times \frac{\sigma}{\sqrt{n}}
\\
\\0.242\pm 1.96 \times (\frac{0.037}{\sqrt{10}})
\\
\\0.242 \pm .0229
\\
\\0.242-0.0229=0.2191
\\
\\0.242+0.0229 = 0.2649
5 0
3 years ago
An airplane travels 6111 kilometers against the wind in 9 hours and 7911 kilometers with the wind in the same amount of time. Wh
natima [27]

Answer:

Speed of the plane in still air: 779\; {\rm km \cdot h^{-1}}.

Windspeed: 100\; {\rm km \cdot h^{-1}}.

Step-by-step explanation:

Assume that x\; {\rm km \cdot h^{-1}} is the speed of the plane in still air, and that y\; {\rm km \cdot h^{-1}} is the speed of the wind.

  • When the plane is travelling against wind, the ground speed of this plane (speed of the plane relative to the ground) would be (x - y)\; {\rm km \cdot h^{-1}}.
  • When this plane is travelling in the same direction as the wind, the ground speed of this plane would be (x + y)\; {\rm km \cdot h^{-1}}.

The question states that when going against the wind (v = (x - y)\; {\rm km \cdot h^{-1}},) the plane travels 6111\; {\rm km} in 9\; {\rm h}. Hence, 9\, (x - y) = 6111.

Similarly, since the plane travels 7911\; {\rm km} in 9\; {\rm h} when travelling in the same direction as the wind (v = (x + y)\; {\rm km \cdot h^{-1}},) 9\, (x + y) = 7911.

Add the two equations to eliminate y. Subtract the second equation from the first to eliminate x. Solve this system of equations for x and y: x = 779 and y = 100.

Hence, the speed of this plane in still air would be 779\; {\rm km \cdot h^{-1}}, whereas the speed of the wind would be 100\; {\rm km \cdot h^{-1}}.

3 0
1 year ago
On a piece of paper, show how to find the slope between these two points: (-1,5) and (3,13). Show all of your work and simplify
dybincka [34]
Use the formula m = y2 - y1
————
x2 - x1
(-1 is x1, 5 is y1) (3 is x2, 13 is y2)
13 - 5 8
——— = — = 2
3 - (-1) 4

2 is the slope
6 0
3 years ago
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