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Alex777 [14]
3 years ago
14

1. What is the average of 12, 43, 51, 62? A 32 B51 C42 D 36 NEED HELP!

Mathematics
2 answers:
mixas84 [53]3 years ago
6 0

Answer:

42

Step-by-step explanation:

Average =

Sum of the terms/ number of terms

Terms here refers to the data or the values given

So,

The sum is 12+43+51+62= 168

The number of terms is 4

Therefore the average is,

168/4

= 42

mario62 [17]3 years ago
5 0

Answer:

42

Step-by-step explanation:

Average =  

Sum

/Count

=  

168/4

=  42

MARK ME BRAINLIEST

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Find the value of x <br> A) 6.0<br> B) 7.7<br> C) 60<br> D) 12
galben [10]

Answer:

D 12

Step-by-step explanation:

The product of the lengths of the segments of one chord equal the product of the segments of the other.

x*4 = 6*8

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Divide each side by 4

4x/4 = 48/4

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4 0
3 years ago
Select the equivalent expression. <br><br> A, B or C?
Alina [70]

Answer:

  A

Step-by-step explanation:

The relevant rules of exponents are ...

  (a^b)^c = a^(bc)

  a^-b = 1/a^b

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  \left(\dfrac{2^{-10}}{4^2}\right)^7=\dfrac{2^{(-10)(7)}}{4^{(2)(7)}}=\dfrac{2^{-70}}{4^{14}}\\\\=\boxed{\dfrac{1}{2^{70}\cdot 4^{14}}}

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4 years ago
. What is the value of x in the equation –6 + x = –2?
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4 is the right answer
5 0
4 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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