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maxonik [38]
3 years ago
7

Find the measure of an exterior angle of a decagon

Mathematics
1 answer:
miv72 [106K]3 years ago
6 0

\bf \textit{sum of all exterior angles in a polygon}\\\\n\theta =360~~\begin{cases}n=\textit{number of}\\\qquad sides\\\theta =angle~in\\\qquad degrees\\[-0.5em]\hrulefill\\n=10\end{cases}\implies 10\theta =360\implies \theta =\cfrac{360}{10}\implies n=36

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I’ll give brainliest!!
nignag [31]

Answer:

If A is wrong then B I think

Step-by-step explanation:

3 0
3 years ago
Can someone please help me with this question?!? I am so confused and I don't know how to answer it.
lbvjy [14]

9514 1404 393

Answer:

  a. f(0) = 1

  b. DNE (does not exist)

  c. DNE

  d. lim = 3

Step-by-step explanation:

The function exists at a point if it is defined there. The function is defined anywhere on the solid line and at solid dots. It is not defined at open circles. So, the function is defined everywhere except (2, 3), which has an open circle.

The open circle at (0, 4) prevents the function from being doubly-defined at x=0, since it is already defined to be 1 at x=0.

This discussion tells you ...

  f(0) = 1

 f(2) does not exist. There is a "hole" in the function definition there.

__

The function has a limit at a point if approaching from the left and approaching from the right have you approaching that same point.

Consider the point (1, 2). The graph is a solid line through that point. Approaching from values less than x=1, we get to the same point (1, 2) as when we approach from values greater than x=1.

Similarly, consider the point (2, 3). Approaching from values of x less than 2, we get to the same point (2, 3) as when we approach from x-values greater than 2. The limit at x=2 is 3. The only difference from the previous case is that the function is not actually defined to be that value there.

__

Now consider what happens at x=0. When we approach from the left, we approach the point (0, 4). When we approach from the right, we approach the point (0, 1). These are different points. Because they are different coming from the left and from the right, we say "the limit as x→0 does not exist."

__

In summary, ...

  a) f(0) = 1

  b) lim x → 0 does not exist

  c) f(2) does not exist

  d) lim x → 2 = 3

_____

<em>Additional comment</em>

The significance of the function not being defined at a point where the limit exists, (2, 3), is that <em>the function is not continuous there</em>. This kind of discontinuity is called "removable", because we could make the function continuous at x=2 by defining f(2) = 3 (that is, "filling the hole").

6 0
2 years ago
PLEASEEE HELPPPP IM BEGGIN U
Anna71 [15]

Answer:

  a) positive real zeros: 2 or 0; negative real zeros: 2 or 0; complex zeros: 0, 2, or 4. (rule of signs)

  b) ∪-shaped, as for an even-degree polynomial with positive leading coefficient. See attached.

  c, d) See attached

Step-by-step explanation:

Descarte's rule of signs gives bounds on the number of positive and negative real roots. The numbers it gives may be reduced by multiples of 2, as complex roots will come in conjugate pairs. The total number of roots of all kinds will match the degree of the polynomial.

Synthetic division is essentially polynomial long division with some modifications:

  • the variables are omitted ("place value" is used instead)
  • the constant in the divisor is <em>negated</em> so its product with the partial quotient can be <em>added</em> to obtain the new dividend
  • the divisor binomial is assumed to have a leading coefficient of 1.

__

<h3>a) </h3>

The signs of the terms of the given polynomial are + - - - +. There are two sign changes, so 2 possible positive real roots.

When the signs of the odd-degree terms are changed, the signs become + + - + +. There are still two sign changes, so 2 possible negative real roots.

Either or both of these numbers can be reduced by 2 if the roots include a conjugate pair. That is, there may also be 0 possible positive real roots, and 0 possible negative real roots.

The number of non-real (complex) zeros may be any multiple of 2 up to the degree of the polynomial. The may be 0, 2, or 4 possible non-real zeros.

__

<h3>b)</h3>

The graph is the first attachment. It shows 4 real zeros: x = -4, -2, 1, 6.

Since the polynomial is of even degree (4) and has a positive leading coefficient (+1), we expect the general shape to be ∪-shaped. It is.

__

<h3>c)</h3>

The second attachment shows synthetic division using x = -4. (The binomial divisor is (x+4).) The remainder (lower right value in the tableau) is the value of y when x=-4. The third attachment shows synthetic division using the value x=3. (y is -210 when x=3.)

Maybe this is the table you want:

  \begin{array}{|c|c|c|}\cline{1-3}x&-4&3\\\cline{1-3}y&0&-210\\\cline{1-3}\end{array}

__

<h3>d)</h3>

The second attachment shows synthetic division by the factor (x+4).

_____

<em>Additional comment</em>

The synthetic division attachments show instructions for carrying out the synthetic division and interpreting the results. As we said above, "the entry on the left" is the opposite of the constant in the binomial divisor. It is the actual value of x you want to use to evaluate the function.

The equations shown are merely for the purpose of indicating the operations that are used. An actual synthetic division tableau is simply a 3-row table of numbers, with the bottom row being the quotient coefficients and remainder.

6 0
2 years ago
A teacher weighed 145 lbs in 1986 and weighs 190 lbs in 2007. What is the rate of change in weight? HELP
marishachu [46]

Answer:

c

Step-by-step explanation:

she gained 45 pounds which is 15/7 of 145

8 0
3 years ago
Read 2 more answers
Dale finished his English assignment in 1/3 hours . Then he completed his chemistry assignment in 1/2 hours . How much more time
11Alexandr11 [23.1K]

Answer:

1/6 hours

Step-by-step explanation:

Since it wants a fraction form all you have to do is subract 1/3 from 1/2

1/2 - 1/3

cross multiply:

3/2-2/3

subtract the numerator but multiply the denominator and simplify if you have to:

1/6

6 0
3 years ago
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