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Kobotan [32]
1 year ago
11

Solve for x. leave your answer in simplest radical form.

Mathematics
1 answer:
adell [148]1 year ago
8 0

Answer:

could you provide the equation for which x is to be solved?:)...If your question was supposed to be that ...then hope the solution helps:)

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Find m∠FHG. Hurry pls
sesenic [268]
180 -20 = 160 so angle FHG equals 160
8 0
2 years ago
A pound of rice crackers costs $2.88. Matthew purchased 1/4 pound of crackers. How much did he pay for crackers?
Rom4ik [11]
He payed 0.72 
First you take 2.88, then you multiply it by 1/4 or a simpler way would be to divide 2.88 by 4. The answer is $0.72
6 0
3 years ago
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If you actually answer the question and do not give me some bogus answer(ex: I do not know) I will mark as brainliest to whoever
7nadin3 [17]

Answer:

I cannot access your picture

Step-by-step explanation:

8 0
3 years ago
Reason Can you subtract a positive integer from a positive integer
Afina-wow [57]

Answer:

No

Step-by-step explanation:

No matter the situation, when you multiply a negative by a negativeyou get a positive and a positive by a positive you get a positive. but if its two different like a negative and a positive then its NEGITIVE.

let's say you have 23 and you're multiplying by 2.

It's always increasing so it doesnt ever reach the negitive numbers.

6 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
2 years ago
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