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tatuchka [14]
3 years ago
6

Show that (x+1) is a factor of f(x)=19x^42+18x-1

Mathematics
1 answer:
Dmitry_Shevchenko [17]3 years ago
6 0

Answer:

see explanation

Step-by-step explanation:

If (x + 1) is a factor then f(- 1) = 0

f(- 1) = 19(-1)^{42} + 18(- 1) - 1 = 19 - 18 - 1 = 0

Thus (x + 1) is a factor of f(x)

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Need help ASAP. i forgot all the steps after the Area is solved with 5. or if 25pi is the right answer.
maxonik [38]

Answer:

42.5\:\mathrm{m^2}

Step-by-step explanation:

The area of a sector with measure \theta in degrees is given by r^2\pi\cdot\frac{\theta}{360}, where r is the radius of the sector.

What we're given:

  • r of 5
  • \theta of 195^{\circ}

Solving, we get:

A_{sector}=5^2\pi\cdot \frac{195}{360}=13.5416666667\pi=42.5424005174\approx \boxed{42.5\:\mathrm{m^2}}

*Notes:

  • units should be in square meters (area)
  • the problem does not say whether to round or leave answers in term of pi, so you may need to adjust the answer depending on what your teacher specifically wants
7 0
3 years ago
Determine whether the system of equations below has one solution, infinitely many
Iteru [2.4K]
No solution because I know I’m right
6 0
3 years ago
The reciprocal of 5 plus the reciprocal of 7 is the reciprocal of what number?
Natalija [7]
2/5 is the answer so the reciprocal would be 5/2
4 0
3 years ago
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Which statements are true about the interquartile range? Select all that apply. Subtract the lowest and highest values to find t
katen-ka-za [31]

Answer:

The correct options are:

Interquartile ranges are not significantly impacted by outliers.

Lower and upper quartiles are needed to find the interquartile range.

The data values should be listed in order before trying to find the interquartile range.


The option Subtract the lowest and highest values to find the interquartile range is incorrect because the difference between lowest and highest values will give us range.

The option A small interquartile range means the data is spread far away from the median is incorrect because a small interquartile means data is nor spread far away from the median

6 0
3 years ago
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A square is inscribed in a circle. If the area of the square is 9 in2, what is the ratio of the circumference of the circle to t
IgorC [24]

The ratio of the circumference of the circle to the perimeter of the square is \pi \sqrt{2}:4

<em><u>Explanation</u></em>

Area of the square = 9 inch²

If the side length of the square is a, then

a^2 = 9 \\ \\ a= \sqrt{9} =3

So the side length of the square is 3 inch.

Now as the square is inscribed in a circle, so the diagonal of the square will be diameter of the circle.

Length of the diagonal of square = a\sqrt{2} = 3\sqrt{2} inch

So, the diameter of the circle = 3\sqrt{2} inch

If the radius of the circle is r, then

2r = 3\sqrt{2} \\ \\ r= \frac{3\sqrt{2}}{2}

Circumference of the circle, 2\pi r= 2\pi *\frac{3\sqrt{2}}{2}= 3\pi \sqrt{2} inch

and Perimeter of the square, 4a = (4*3)inch= 12 inch

So, the ratio will be: 3\pi \sqrt{2} : 12 = \pi \sqrt{2} : 4

8 0
3 years ago
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