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KATRIN_1 [288]
3 years ago
11

Someone Help me solve these by substitution please !

Mathematics
1 answer:
PSYCHO15rus [73]3 years ago
8 0
I think the answer is:

X=3

Y=-2
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PLEASE I NEED HELP ​
NeTakaya

Answer:

D) y=-1/2x+4

Step-by-step explanation:

D) y=-1/2x+4

In order to be parallel, it has to have the same slope (-1/2).  Also, if you plug in (-2,5) for x and y, it does pass through those coordinates.

8 0
3 years ago
If 11 - 3x = x, then 20x=
dimaraw [331]

Answer:

20x = 55

Step-by-step explanation:

First you have to add 3x to both sides of 11-3x=x

11-3x=x\\+3x=+3x\\11=4x

Now that you have 11 = 4x you divide 4 on both sides \frac{11}{4} =\frac{4x}{4}

11/4 = x

But you can't divide 11 by 4 unless you are looking for the decimal form of 11/4 which is 2.75

so now we take 11/4 or 2.75 depending on if you need your answer in fraction or as a decimal and replace that with x  however both will equal the number 55

20(\frac{11}{4})=55\\20(2.75)=55

4 0
3 years ago
Evaluate the Expression :<br> (-2 4/5) + 2 2/3 + (-3 1/2)
denis-greek [22]

−24/5+22/3+−31/2

=38/15+−31/2

=−389/30

7 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
The cross-sectional areas of a triangular prism and a right cylinder are congruent. The triangular prism has a height of 5 units
Fiesta28 [93]

u go to central too lol

6 0
2 years ago
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