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bearhunter [10]
2 years ago
8

THIS IS NOT A QUESTION! THE PPL THAT ANSWERED THIS QUESTION WAS WRONG!!

Mathematics
2 answers:
natka813 [3]2 years ago
6 0

Ok, good job for posting the answer

ankoles [38]2 years ago
3 0

correct

90 and 24

follow me on big brain

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12 over 16, 30 over 40
UNO [17]
In fraction is 3/4 and 3/4
Decimal is .75 and .75
4 0
3 years ago
8x^2+5=22x Need help learning factoring please put work so I can understand
sp2606 [1]

Answer: (2x - 5) • (4x - 1)

STEP 1 : Equation at the end of step 1

(23x2 - 22x) + 5

STEP 2 : Trying to factor by splitting the middle term

2.1 Factoring 8x2-22x+5

The first term is, 8x2 its coefficient is 8 .

The middle term is, -22x its coefficient is -22 .

The last term, "the constant", is +5

Step-1 : Multiply the coefficient of the first term by the constant 8 • 5 = 40

Step-2 : Find two factors of 40 whose sum equals the coefficient of the middle term, which is -22 .

-40 + -1 = -41

-20 + -2 = -22 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -20 and -2

8x2 - 20x - 2x - 5

Step-4 : Add up the first 2 terms, pulling out like factors :

4x • (2x-5)

Add up the last 2 terms, pulling out common factors :

1 • (2x-5)

Step-5 : Add up the four terms of step 4 :

(4x-1) • (2x-5)

Which is the desired factorization

Final result : (2x - 5) • (4x - 1)

5 0
3 years ago
A. 15<br> b. 16<br> c. 9<br> d. 14
MAXImum [283]
A. 15
0 +0=0
0 +1 = 1
1+2=3
3+3=6
6+4=10
10+5 = 15
3 0
3 years ago
Read 2 more answers
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
HI I NEED HELP IVE BEEN DOING HOMEWORK FOR 6 HOURS ummmmm anyways, how do i express this (y^3•y^6 - in picture) in exponential f
LUCKY_DIMON [66]

y^3 = y * y * y

y^6 = y * y * y * y * y * y

6 0
3 years ago
Read 2 more answers
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