MATHEMATICS HIGH SCHOOL

A sign in a store window has the shape of a regular nonagon (9-sided figure). The sign has an apothem length of 40 cm and a side length of 29 cm.What is its area?

____cm2

Answers

Answer 1
Answer: Area = A
Side length = L = 29 cm
number of sides = n = 9
Apothen = a = 40 cm

A = nLa / 2 = 9*29cm*40cm / 2 = 5,220 cm^2

Answer: 5,220 cm^2
Answer 2
Answer:

Answer:

5220 cm

Step-by-step explanation:

A=9/4(a^2)cot(180°/9) is the formula and got 5,220 cm

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Find the supplement of the given angle. Then draw the two angles as linear pairs. Label each angle with its measure.m angle ABC=72 B will be the vertex

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The supplement angle of angle ABC is 108 degrees because, supplement angles mean angles that add up each other to 180 degrees so 180-72=108

The distance from town A to town B is five miles. C is six miles from B. Which of the following could be the distance from A to C?I 11
II 1
III 7

Answers

It's 11 because you add the 5 from a to b and b to c

Find the range of the following piecewise function.[0,11)
(-4,6]
(0,11]
[-4,6)

Answers

Given:

The piecewise function is

f(x)=\begin{cases}x+4 & \text{ if } -4\leq x<3 \n 2x-1 & \text{ if } 3\leq x<6 \end{cases}

To find:

The range of given piecewise function.

Solution:

Range is the set of output values.

Both functions f(x)=x+4 and f(x)=2x-1 as linear functions.

Starting value of f(x)=x+4 is at x=-4 and end value is at x=3.

Starting value: f(-4)=-4+4=0

End value: f(3)=3+4=7

Starting value of f(x)=2x-1 is at x=3 and end value is at x=6.

Starting value: f(3)=2(3)-1=5

End value: f(6)=2(6)-1=11

Least range value is 0 at x=-4 and 0 is included in the range because -4 is included in the domain.

Largest range value is 11 at x=6 and 11 is not included in the range because 6 is not included in the domain.

So, the range of the given piecewise function is [0,11).

Therefore, the correct option is A.

Is this correct ? .......................

Answers

To find the midpoint of the two points (x₁ ,y₁) and (x₂, y₂) we need to use the formula:

\sf{Midpoint=((x_1 + x_2)/(2), (y_1+y_2)/(2))}

So to find the midpoint of E (a,a) and F (3a, a), let's plug it in to the formula:

\sf{Midpoint=((a + 3a)/(2), (a+a)/(2))}

Simplify the numerator:

\sf{Midpoint=((4a)/(2), (2a)/(2))}

Simplifying the fractions more:

\sf{Midpoint=(2a, a)}


So the midpoint of EF is (2a,a).

Your answer of (a,a) would be wrong.

Which graph represents the solution to the system of inequalities?x + y ≤ 4

y – x ≥ 1

Answers

Given inequalities are

x+y\le4 and

y-x\ge1

Now we will graph both inequalities to get the common region which represents solution.

Work for x+y\le4

first we graph the line x+y=4 then shade the graph for inequality symbol \le

we can plug any random number for x and find y-value to get points

lets plug x=0 then we get:

x+y=4

0+y=4

y=4

so the point is (0,4)

Similarly we can find more point like (4,0).

Now graph both points and join both points by straight line.

Now we find direction of shading.

pick any test point which is not on that line x+y=4 say (0,0)

and plug it into orinal inequality x+y\le4

0+0\le4

0\le4

which is true.

True means shade in direction of test point. Otherwise we shade in opposite direction.

We will repeat same process for other inequality to graph that.

Hence final graph for the answer will be as shown in attached image.

Triangular region ABC is final answer.

Solve: ∫e^u/(1−eu)^2​du

Answers

Answer:

Hi,

Step-by-step explanation:

Let's\ say\ v=1-e^u \n\n\displaystyle \int\limits {\frac{e^u } { (1-e^u)^2 }} \, du \n\n\n=\int\limits {(1)/(v^2) } \, dv \ \n\n=-(1)/(u) +C\n\n=-\frac{1 } { 1-e^u } + C
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