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beks73 [17]
2 years ago
7

Find the perimeter of the figure to the nearest hundredth.​

Mathematics
1 answer:
WITCHER [35]2 years ago
6 0
Perimeter: 36.84 feet

Let’s start with the easy part...all the rectangle pieces 6 x 3 = 18 feet

We know from subtracting the rectangle length (6ft) from the total length (12 ft) that the diameter of the circle is 6ft.

Circumference = diameter x pi
= 6 x 3.14 = 18.84

We have the circle split into 2 halves, but the circumference remains the same

18 + 18.84 = 36.84
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Determine the value of x for the diagram shown.
TEA [102]

Answer:

C) 60

Step-by-step explanation:

180 - 120

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In the expression below, a is an integer. 12^3+ax-20 if 3x+4 is a factor of the expession above, what is the value of a
SVEN [57.7K]

Our function is:

f(x)=12^3+ax-20

Now, if 3x+4 is a factor of f(x) then x=\frac{-4}{3} must satisfy the equation:

So, f(\frac{-4}{3} )=0

Putting x=\frac{-4}{3}

We get,

f(\frac{-4}{3} )=12^3+a(\frac{-4}{3} )-20=0

12^3+\frac{-4}{3} a-20=0

\frac{-4}{3} a=20-12^3=20-1708

a=-1708 \times \frac{-3}{4} =1281

So, the value of 'a' for the given expression is 1281.

4 0
3 years ago
A movie theater has 200 of its 250 seats occupied . Write the fraction that shows the number of empty seats
guajiro [1.7K]
1/5. Reduce your fraction of 200/250.
6 0
3 years ago
Which of the following is closest to the area of the shaded region below?
enot [183]

Answer:

C is the answer

Step-by-step explanation:

Find the area of the whole circle, r=12cm

=pie× radius sq

=3.14×12×12

=<u>452.16cm sq</u>

The are of the smaller circle, r=6cm

=same formula

=3.14×6×6

=<u>113.04cm sq</u>

<u>Subtract</u><u> </u><u>them</u><u> </u><u>to</u><u> </u><u>find</u><u> </u><u>the</u><u> </u><u>area</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>shaded</u><u> </u><u>region</u><u> </u>

452.16 - 113.04

=339.12

<em>Closest</em><em> </em><em>is</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>339</em><em> </em><em>cm</em><em> </em><em>sq</em><em> </em><em>which</em><em> </em><em>is</em><em> </em><em>c</em>

<em>Hope</em><em> </em><em>it's</em><em> </em><em>helpful</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em> </em>

5 0
3 years ago
Please determine whether the set S = x^2 + 3x + 1, 2x^2 + x - 1, 4.c is a basis for P2. Please explain and show all work. It is
ohaa [14]

The vectors in S form a basis of P_2 if they are mutually linearly independent and span P_2.

To check for independence, we can compute the Wronskian determinant:

\begin{vmatrix}x^2+3x+1&2x^2+x-1&4\\2x+3&4x+1&0\\2&4&0\end{vmatrix}=4\begin{vmatrix}2x+3&4x+1\\2&4\end{vmatrix}=40\neq0

The determinant is non-zero, so the vectors are indeed independent.

To check if they span P_2, you need to show that any vector in P_2 can be expressed as a linear combination of the vectors in S. We can write an arbitrary vector in P_2 as

p=ax^2+bx+c

Then we need to show that there is always some choice of scalars k_1,k_2,k_3 such that

k_1(x^2+3x+1)+k_2(2x^2+x-1)+k_34=p

This is equivalent to solving

(k_1+2k_2)x^2+(3k_1+k_2)x+(k_1-k_2+4k_3)=ax^2+bx+c

or the system (in matrix form)

\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}

This has a solution if the coefficient matrix on the left is invertible. It is, because

\begin{vmatrix}1&1&0\\3&1&0\\1&-1&4\end{vmatrix}=4\begin{vmatrix}1&2\\3&1\end{vmatrix}=-20\neq0

(that is, the coefficient matrix is not singular, so an inverse exists)

Compute the inverse any way you like; you should get

\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}=-\dfrac1{20}\begin{bmatrix}4&-8&0\\-12&4&0\\-4&3&-5\end{bmatrix}

Then

\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}\begin{bmatrix}a\\b\\c\end{bmatrix}

\implies k_1=\dfrac{2b-a}5,k_2=\dfrac{3a-b}5,k_3=\dfrac{4a-3b+5c}{20}

A solution exists for any choice of a,b,c, so the vectors in S indeed span P_2.

The vectors in S are independent and span P_2, so S forms a basis of P_2.

5 0
3 years ago
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