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3 years ago

What is the sum of the interior angles in a hendecagon

Mathematics

2 answers:

murzikaleks [220]3 years ago

5 0

11 sides of the hendecagon.

spin [16.1K]3 years ago

4 0

A hendecagon is an 11 sided polygon. Each internal angle of a regular hendecagon = 147.27 o . The sum of interior angles of a hendecagon = (n - 2) * 180o = (11 - 2) 180o = 1,620o.

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Which equation represents a parabola that opens upward has a minimum at x=3 and has a line of symmetry at x=3

umka2103 [35]

Answer:

$y=x^2-6x+5$

Step-by-step explanation:

Let us consider the equation $y=x^2-6x+5$

For a quadratic equation in a standard form, $y=ax^2+bx+c$ , the axis of symmetry is the vertical line $x = \frac{-b}{2a}$ .

Here in this case we have, $a=1, b=-6 , c =5$

Putting the values we get,

$x = \frac{-(-6)}{2\times 1} = \frac{6}{2} =3$

We can see that the axis of symmetry is x=3 and the graph is giving minimum at x=3.

Therefore, the required equation is $y=x^2-6x+5$ . Refer the image attached.

4 0

3 years ago

Read 2 more answers

DO NOT SKIP!! Let’s see who’s smart enough to work out the problem if u get it correct I’ll mark u as BRAINLIEST SHOW THE WORK P

forsale [732]

Answer:

y=10.1(x)

x= $\frac{y}{10.1}$

Step-by-step explanation:

It is said in the problem that y would represent how much money she earns and x represents the number hours she works.

so it also said that the amount of Jessa earns is proportional to the amount of time she works .

finding y will be easy dividing 252.50 by 25 will give you 10.1

so how did we get that equation

y will represent the amount and x will represent the hours

4 0

3 years ago

Consider the original complex figure and the reduction.

Dmitriy789 [7]

Answer:

1/8

Step-by-step explanation:

I'm going to try to explain this as easy as possible. What I did was take the original shape and divide it by the new shape. For this question, I solved it by dividing 32(the original base) by 4(the new base) and got 8. So the scale factor of the reduction was 1/8.

8 0

3 years ago

How to solve this trig

n200080 [17]

Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

First Question

What we have to do is to isolate cos first.

$\displaystyle \large{ cos \theta = - \frac{1}{2} }$

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

Find Q2

$\displaystyle \large{ \pi - \frac{ \pi}{3} = \frac{3 \pi}{3} - \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{2 \pi}{3} }$

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{3} = \frac{3 \pi}{3} + \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{4 \pi}{3} }$

Both values are apart of the interval. Hence,

$\displaystyle \large \boxed{ \theta = \frac{2 \pi}{3} , \frac{4 \pi}{3} }$

Second Question

Isolate sin(4 theta).

$\displaystyle \large{sin 4 \theta = - \frac{1}{ \sqrt{2} } }$

Rationalize the denominator.

$\displaystyle \large{sin4 \theta = - \frac{ \sqrt{2} }{2} }$

The problem here is 4 beside theta. What we are going to do is to expand the interval.

$\displaystyle \large{0 \leqslant \theta < 2 \pi}$

Multiply whole by 4.

$\displaystyle \large{0 \times 4 \leqslant \theta \times 4 < 2 \pi \times 4} \\ \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}$

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{4} = \frac{ 4 \pi}{4} + \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{5 \pi}{4} }$

Find Q4

$\displaystyle \large{2 \pi - \frac{ \pi}{4} = \frac{8 \pi}{4} - \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{7 \pi}{4} }$

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

$\displaystyle \large{ \theta + 2 \pi k = \theta \: \: \: \: \: \sf{(k \: \: is \: \: integer)}}$

Hence:-

For Q3

$\displaystyle \large{ \frac{5 \pi}{4} + 2 \pi = \frac{13 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 4\pi = \frac{21 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 6\pi = \frac{29 \pi}{4} }$

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

For Q4

$\displaystyle \large{ \frac{ 7 \pi}{4} + 2 \pi = \frac{15 \pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 4 \pi = \frac{23\pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 6 \pi = \frac{31 \pi}{4} }$

Therefore:-

$\displaystyle \large{4 \theta = \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4} }$

Then we divide all these values by 4.

$\displaystyle \large \boxed{\theta = \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16} }$